{"ret_code": 0, "ret_str": "Successful", "tot_pages": 40, "hits": [{"docid": 33619, "score": 9.804, "title": "Indefinite integral of trignometric function", "url": "http://math.stackexchange.com/questions/880315/indefinite-integral-of-trignometric-function", "snippet": "cos x}{\\cos x(1+\\cos x)}dx[\/imath] [imath]=\\int\\frac{2\\sin^2\\dfrac x2}{\\cos^2\\dfrac x2-\\sin^2\\dfrac x2}\\frac12\\sec^2\\dfrac x2dx[\/imath] [imath]=\\int\\frac{2\\tan^2\\dfrac x2}{1-\\tan^2\\dfrac x2}\\frac12\\sec^2\\dfrac x2dx[\/imath] [imath]=\\int\\frac{2u^2-2+2}{1-u^2}du=-2u+2\\int\\frac{du}{1-u^2}du[\/imath] Hint : \\begin{align} \\int<\/em>\\frac<\/em>{1-\\cos<\/em> x<\/em>}{\\cos<\/em>(1+\\cos<\/em> x<\/em>)}\\ dx&=\\int<\/em>\\frac<\/em>{\\color{red}{1+\\cos<\/em> x<\/em>}-2\\cos<\/em> x<\/em>}{\\cos<\/em>(\\color{red}{1+\\cos<\/em> x<\/em>})}\\ dx\\\\ &=\\int<\/em>\\frac<\/em>1{\\cos<\/em> x<\/em>}\\ dx-\\int<\/em>\\frac<\/em>{2}{1+\\cos x<\/em>}\\ dx\\\\ &=\\int<\/em>\\frac{\\cos x}{\\cos^2 x}\\ dx-\\int\\frac{2}{1+\\cos x}\\cdot\\frac{1-\\cos x}{1-\\cos x}\\ dx\\\\ &=\\int\\frac{d(\\sin x)}{1-\\sin^2 x}-\\int\\frac{2-2\\cos x}{1-\\cos^2 x}\\ dx\\\\ &=\\int\\frac{d(\\sin x)}{2(1-\\sin x)}-\\int\\frac{d(\\sin x)}{2(1+\\sin x)}-\\int\\frac{2-2\\cos x}{\\sin^2 x}\\ dx\\\\ &=\\int\\frac{d(\\sin x)}{2(1-\\sin x)}-\\int\\frac{d(\\si ... "}, {"docid": 55167, "score": 9.453, "title": "Riemann Stieltjes Integral with trig limit of integration!", "url": "http://math.stackexchange.com/questions/648534/riemann-stieltjes-integral-with-trig-limit-of-integration", "snippet": "0}^{\\frac{\\pi}{4}}|\\cos2x|d(\\sin x)+ \\int_{\\frac{\\pi}{4}}^{\\frac{-\\pi}{2}}|\\cos2x|d(\\sin x)[\/imath] [imath]+\\int_{\\frac{-\\pi}{2}}^{\\frac{-\\pi}{4}}|\\cos2x|d(\\sin x)+\\int_{\\frac{-\\pi}{4}}^{0}|\\cos2x|d(\\sin x)[\/imath] pls going forward from here, how do i get rid of the modulus signs. thanks alot. Hints:\\begin{align*} \\int<\/em>_{\\frac<\/em>{-\\pi}{2}}^{\\frac<\/em>{\\pi}{2}}|\\cos<\/em>2x<\/em>|d(\\sin<\/em> x<\/em>)&=2\\int<\/em>_{0}^{\\frac<\/em>{\\pi}{2}}|\\cos<\/em>2x<\/em>|d(\\sin<\/em> x<\/em>)\\\\ &=2\\int<\/em>_{0}^{\\frac<\/em>{\\pi}{4}}|\\cos<\/em>2x<\/em>|d(\\sin<\/em> x<\/em>)+2\\int<\/em>_{\\frac<\/em>\\pi4}^{\\frac<\/em>{\\pi}{2}}|\\cos<\/em>2x<\/em>|d(\\sin x)\\\\ &=2\\int_{0}^{\\frac{\\pi}{2}}\\cos2xd(\\sin x)-2\\int_{\\frac\\pi4}^{\\frac{\\pi}{2}}\\cos2xd(\\sin x) \\end{align*} thank u. But can u do me a favour by explaining this pls, i really want to know the techniques in case of other problems like this nature. do u think this [imath]-\\frac{3\\sqrt{2}}{2}[\/imath] is the ans ... "}, {"docid": 78079, "score": 9.401, "title": "Evaluation of [imath] \\int\\frac{\\sin (x+\\alpha)}{\\cos^3 x}\\cdot \\sqrt{\\frac{\\csc x+\\sec x}{\\csc x-\\sec x}}dx[\/imath]", "url": "http://math.stackexchange.com/questions/853888/evaluation-of-int-frac-sin-x-alpha-cos3-x-cdot-sqrt-frac-csc-x-se", "snippet": " \\alpha\\right)\\cdot \\sqrt{\\frac{1+\\tan x}{1-\\tan x}}\\cdot \\sec^2 xdx[\/imath] Now Let [imath]\\tan x= t\\;,[\/imath] Then [imath]\\sec^2 xdx = dt[\/imath] [imath]\\displaystyle I = \\int (\\cos \\alpha\\cdot t + \\sin \\alpha)\\sqrt{\\frac{1+t}{1-t}}dt[\/imath] Now How can I solve after that Help me Thanks Let \\begin{align} I = \\int<\/em> \\frac<\/em>{\\sin<\/em>(x<\/em>+a)}{\\cos<\/em>^3(x<\/em>)}\\cdot \\sqrt{\\frac<\/em>{\\cos<\/em> x<\/em>+\\sin<\/em> x<\/em>}{\\cos<\/em> x<\/em>-\\sin<\/em> x<\/em>}}dx \\end{align} then \\begin{align} I &= \\int<\/em> (\\sin<\/em>(x<\/em>) \\cos<\/em>(a) + \\cos<\/em>(x<\/em>) \\sin<\/em>(a))(\\cos<\/em>(x) + \\sin<\/em>(x)) \\ \\frac{dx}{\\cos^{3}(x) \\sqrt{\\cos(2x)}} \\\\ &= \\cos(a) \\ \\int \\frac{\\sin^{2}(x) \\ dx}{\\cos^{3}(x) \\sqrt{\\cos(2x)}} + \\sin(a) \\ \\int \\frac{ dx}{\\cos(x) \\sqrt{\\cos(2x)}} + (\\cos(a) + \\sin(a)) \\ \\int \\frac{\\sin(x) \\ dx}{\\cos^{2}(x) \\sqrt{\\cos(2x)}} \\\\ &= \\frac{1}{2} (\\cos(a) + 2 \\sin(a)) \\ \\tan^{-1}\\left( \\frac{\\si ... "}, {"docid": 255986, "score": 9.374, "title": "Computing [imath]\\int_0^{2\\pi}\\sqrt{1+\\sin x}dx[\/imath]", "url": "http://math.stackexchange.com/questions/1697421/computing-int-02-pi-sqrt1-sin-xdx", "snippet": "Computing [imath]\\int_0^{2\\pi}\\sqrt{1+\\sin x}dx[\/imath] Problem Evaluate the following integral [imath]\\int_0^{2*\\pi}\\sqrt{1+\\sin x}dx[\/imath] Attempted solution Note that \\begin{align*} & \\int<\/em>_{0}^{2\\pi}\\sqrt{1+\\sin<\/em> x<\/em>}dx\\\\ = & \\int<\/em>_{0}^{\\pi}\\sqrt{1+\\sin<\/em> x<\/em>}dx+\\int<\/em>_{\\pi}^{2\\pi}\\sqrt{1+\\sin<\/em> x<\/em>}dx\\\\ = & \\int<\/em>_{0}^{\\pi}\\sqrt{\\left(\\cos<\/em>\\frac<\/em>{x<\/em>}{2}+\\sin<\/em>\\frac<\/em>{x<\/em>}{2}\\right)^{2}}dx+\\int<\/em>_{0}^{\\pi}\\sqrt{1+\\sin<\/em>\\left(x<\/em>-\\pi\\right)}d\\left(x<\/em>-\\pi\\right)\\\\ = & \\int<\/em>_{0}^{\\pi}\\sqrt{\\left(\\cos<\/em>\\frac<\/em>{x<\/em>}{2}+\\sin\\frac{x}{2}\\right)^{2}}dx+\\int_{0}^{\\pi}\\sqrt{1-\\sin x}dx\\\\ = & \\int_{0}^{\\pi}\\sqrt{\\left(\\cos\\frac{x}{2}+\\sin\\frac{x}{2}\\right)^{2}}dx+\\int_{0}^{\\pi}\\sqrt{\\left(\\cos\\frac{x}{2}-\\sin\\frac{x}{2}\\right)^{2}}dx\\\\ = & \\int_{0}^{\\pi}\\left|\\cos\\frac{x}{2}+\\sin\\frac{x}{2}\\right|dx+\\int_{0}^{\\pi}\\left|\\cos\\frac{x}{2} ... "}, {"docid": 177555, "score": 9.331, "title": "definite integral [imath]\\int_{0}^{\\frac{\\pi}{4}} \\frac{\\sin^2x\\cos^2x}{\\sin^3x+\\cos^3x}dx[\/imath]", "url": "http://math.stackexchange.com/questions/2921300/definite-integral-int-0-frac-pi4-frac-sin2x-cos2x-sin3x-cos3", "snippet": "\\displaystyle \\int^{\\frac{\\pi}{4}}_{0}\\frac{\\sin ^2 x \\cos^2 x}{(\\sin^3 x+\\cos^3 x)^2}dx.[\/imath] Using identities [imath]\\sin x+\\cos x=\\sqrt{2}\\cos(x-\\dfrac{\\pi}{4})[\/imath] [imath]\\sin x\\cos x=\\cos^2(x-\\dfrac{\\pi}{4})-\\dfrac12[\/imath] and then substitution [imath]\\dfrac{\\pi}{4}-x=u[\/imath] gives \\begin{align} I &= \\int<\/em>_{0}^{\\frac<\/em>{\\pi}{4}} \\frac<\/em>{\\sin<\/em>^2x<\/em>\\cos<\/em>^2x<\/em>}{\\sin<\/em>^3x<\/em>+\\cos<\/em>^3x<\/em>}dx \\\\ &= \\int<\/em>_{0}^{\\frac<\/em>{\\pi}{4}}\\frac<\/em>{\\sin<\/em>^2x<\/em>\\cos<\/em>^2x<\/em>}{(\\sin<\/em> x<\/em>+\\cos<\/em> x<\/em>)(1-\\sin<\/em> x\\cos<\/em> x)}dx \\\\ &= \\dfrac{1}{2\\sqrt{2}}\\int_{0}^{\\frac{\\pi}{4}}\\frac{(2\\cos^2u-1)^2}{\\cos u(3-2\\cos^2u)}\\ du \\\\ &= \\dfrac{1}{2\\sqrt{2}}\\int_{0}^{\\frac{\\pi}{4}}\\frac{(2\\cos^2u-1)^2}{\\cos^2u(3-2\\cos^2u)}\\cos u\\ du \\end{align} now let [imath]\\sin u=w[\/imath] therefore \\begin{align} I &= \\dfrac{1}{2\\sqrt{2}}\\int_{0}^{\\frac{\\sq ... "}, {"docid": 80294, "score": 9.327, "title": "Calculation of [imath]\\int \\sin^2 x\\frac{\\sin x-\\cos x}{\\sin x+\\cos x}dx[\/imath]", "url": "http://math.stackexchange.com/questions/2529631/calculation-of-int-sin2-x-frac-sin-x-cos-x-sin-x-cos-xdx", "snippet": "in x+\\cos x}dx[\/imath] For (b) Can we solve it any geometrical way like taking unit circle . thanks Shouldn't it be [imath](\\cos x + \\sin x) + (\\cos x - \\sin x)[\/imath] in the first set of brackets since the numerator is [imath]\\sin^2 x \\cos x[\/imath]? what integral do you mean exactly? For (a), \\begin{eqnarray} &&\\int<\/em>\\frac<\/em>{\\sin<\/em>^2 x<\/em>\\cos<\/em> x<\/em>}{\\sin<\/em> x<\/em>+\\cos<\/em> x<\/em>}dx\\\\ &=&\\int<\/em> \\frac<\/em>{\\sin<\/em>^2x<\/em>(\\cos<\/em>^2x<\/em>-\\sin<\/em> x<\/em>\\cos<\/em> x<\/em>)}{\\cos<\/em>^2x-\\sin<\/em>^2x}dx\\\\ &=&\\frac<\/em>14\\int<\/em> \\frac{(1-\\cos(2x))(1+\\cos(2x)-\\sin(2x))}{\\cos(2x)}dx\\\\ &=&\\frac14\\int\\frac{1-\\cos^2(2x)-(1-\\cos(2x))\\sin(2x)}{\\cos(2x)}dx\\\\ &=&\\frac14\\int(\\sec(2x)-\\cos(2x)-\\tan(2x)+\\sin(2x))dx\\\\ &=&\\frac14(\\frac12\\ln(\\frac{\\cos x+\\sin x}{\\cos x-\\sin x})-\\frac12\\sin(2x)+\\frac12\\ln(\\cos (2x))-\\frac12\\cos(2x))+C\\\\ &=&\\frac18(\\ln(\\frac ... "}, {"docid": 246801, "score": 9.255, "title": "Integration by parts and choosing [imath]u[\/imath] and [imath]\\mathrm{d}v[\/imath]", "url": "http://math.stackexchange.com/questions/974033/integration-by-parts-and-choosing-u-and-mathrmdv", "snippet": "ebra. Please, use correct formatting this will enable us to help you Also, please refrain from asking multiple unrelated problems in a single question. put mathematical expressions in [imath] to be readable LaTeX'd. (P.S. I put [\/imath]newcommand\\diff{\\mathop{}\\!\\mathrm{d}}[imath] in the top so I could write [\/imath]\\int<\/em> f(x<\/em>)\\diff x<\/em>[imath] for [\/imath]\\int<\/em> f(x<\/em>)\\diff x<\/em>[imath]. See: math.stackexchange.com\/questions\/957390\/integrate-cos-lnx-dx\/… for the first one. [\/imath]\\newcommand\\diff{\\mathop{}\\!\\mathrm{d}}[imath]For [\/imath]\\int<\/em>\\cos<\/em>(\\ln(x<\/em>))\\diff x<\/em>[imath], use [\/imath]u=\\cos<\/em>(\\ln(x<\/em>))[imath] and [\/imath]\\diff v=\\diff x<\/em>[imath]. Thus, [\/imath]\\diff u=-\\frac<\/em>{\\sin<\/em>(\\ln x)}x\\diff x[imath] and [\/imath]v=x[imath]: [\/imath][imath]\\int\\cos(\\ln(x))\\diff x=uv-\\int v\\diff u=x\\cos(\\ln(x))+\\int\\sin(\\ln(x))\\diff x[\/imath][imath] Integrate by parts again, this time with [\/imath]U=\\sin<\/em>(\\ln(x))[imath] and [\/imath]\\diff V=\\diff x[imath]. Thus, [\/imath]\\diff U=\\frac<\/em>{\\cos<\/em>(\\ln(x))}x[imath] and [\/imath]V=x[imath]: \\begin{align} \\int\\cos(\\ln(x))\\diff x&=x\\cos(\\ln(x))+\\int\\sin(\\ln(x))\\diff x\\\\ &=x\\cos(\\ln(x))+\\left(UV-\\int V\\diff U\\right)\\\\ &=x\\cos(\\ln(x))+\\left(x\\sin(\\ln(x))-\\int\\cos(\\ln(x))\\diff x\\right)\\\\ &=x\\cos(\\ln(x))+x\\sin(\\ln(x))-\\int\\cos(\\ln(x))\\diff x \\end{align} Adding that integral to both sides: \\begin{align} 2\\int\\cos(\\ln(x))\\diff x&=x\\cos(\\ln(x))+x\\sin(\\ln(x))+C\\\\ \\int\\cos(\\ln(x))\\diff x&=\\frac12x\\cos(\\ln(x))+\\frac12x\\sin(\\ln(x))+C \\end{align} For [\/imath]\\int<\/em>\\sec^3x\\diff x[imath], use [\/imath]u=\\sec x[imath] and [\/imath]\\diff v=\\sec^2x\\diff x[imath]. Thus [\/imath]\\diff u=\\tan x\\sec x\\diff x[imath] and [\/imath]v=\\tan x[imath]: [\/imath][imath]\\int\\sec^3x\\diff x=uv-\\int v\\diff u=\\tan x\\sec x-\\int\\tan^2 x\\sec x\\diff x[\/imath][imath] Using [\/imath]\\tan^2 x=\\sec^2 x-1[imath]: \\ ... {align} \\int\\sec^3x\\diff x&=\\tan x\\sec x-\\int(\\sec^2x-1)\\sec x\\diff x\\\\ &=\\tan x\\sec x-\\int(\\sec^3x-\\sec x)\\diff x\\\\ &=\\tan x\\sec x-\\int\\sec^3x\\diff x+\\int\\sec x\\diff x\\\\ &=\\tan x\\sec x-\\int\\sec^3x\\diff x+\\ln|\\tan x+\\sec x|+C \\end{align} where I used the formula for the integral of [\/imath]\\sec[imath]. Adding [\/imath]\\int<\/em>\\sec^3\\diff x[imath] to both sides: \\begin{align} 2\\int\\sec^3x\\diff x&=\\tan x\\sec x+\\ln|\\tan x+\\sec x|+C\\\\ \\int\\sec^3x\\diff x&=\\frac12\\tan x\\sec x+\\frac12\\ln|\\tan x+\\sec x|+C \\end{align} For the third problem, you need to use integration by parts twice. For the last problem, you should probably make a substitution to turn it into [\/imath]\\int<\/em>\\arcsin(x)\\diff x[imath]. Then use [\/imath]u=\\arcsin x[imath] and [\/imath]\\diff v=\\diff x[imath]. I believe your textbook has some integration formulas you could use. The first one is easily solved by letting [\/imath]u=\\ln x[imath] and using the formula for [\/imath][imath]\\int e^{ax} \\cos (bx)\\mathrm{d}x[\/imath][imath] or by integrating by parts with [\/imath]f=\\cos<\/em> u, \\mathrm{d}q=e^u \\mathrm{d}u[imath] The second is the most usual example used on the reduction formula for [\/imath]\\int<\/em> \\sec^m(x) \\mathrm{d}x[imath] Well, the third requires a usual trick(Assuming the integrand is [\/imath]e^{2x} \\sin<\/em> x[imath]), you integrate twice by parts, setting [\/imath]h[imath] the trigonometric, [\/imath]dv[imath] the exponential, your initial integral will reapear somewhere, separate it in LHS, the RHS is easily calculated Last but not least, substitute [\/imath]b=2x[imath] then integrate by parts with [\/imath]j=\\sin^{-1} (x) ... "}, {"docid": 198619, "score": 9.194, "title": "Evaluate [imath]\\int \\cos(3x) \\sin(2x) \\, dx[\/imath].", "url": "http://math.stackexchange.com/questions/703805/evaluate-int-cos3x-sin2x-dx", "snippet": "Evaluate [imath]\\int \\cos(3x) \\sin(2x) \\, dx[\/imath]. Evaluate the indefinite integral \\begin{align} \\int<\/em> \\cos<\/em>(3x<\/em>) \\sin<\/em>(2x<\/em>) \\, dx \\end{align} Please see my work attempt as my answer below. HINT: Using Werner Formulas [imath]2\\cos3x\\sin2x=\\sin(3x+2x)-\\sin(3x-2x)[\/imath] Now use [imath]\\displaystyle\\int\\sin mx\\ dx=-\\frac{\\cos mx}m+C[\/imath] Finally and optionally, we can use Multiple Angle Formula to expand [imath]\\cos5x[\/imath] in the p ... right) - \\frac12 \\left(\\frac{e^{ix}-e^{ix}}{2i}\\right)=\\frac12\\sin 5x - \\frac12\\sin x[\/imath] Then integrate to get [imath]\\int\\cos 3x \\sin 2x \\; dx= \\int\\left(\\frac12\\sin 5x - \\frac12\\sin x\\right)\\;dx= \\boxed{-\\dfrac1{10}\\cos 5x +\\dfrac12\\cos x + C}[\/imath] Employing trigonometric identities, we have \\begin{align} \\int<\/em> \\cos<\/em>3x<\/em> \\sin<\/em>2x<\/em> \\, dx &= \\int<\/em> \\cos<\/em>(2x<\/em>+x<\/em>) \\sin<\/em>2x<\/em> \\, dx \\\\ &= \\int<\/em> [\\cos<\/em>2x<\/em>\\cos<\/em> x - \\sin<\/em>2x\\sin<\/em> x ]\\sin<\/em>2x \\, dx \\\\ &= \\int<\/em> \\cos2x \\cos x \\sin2x - \\sin^2 2x \\sin x \\, dx \\\\ &= \\int (\\cos^2 x - \\sin^2 x) \\cos x (2 \\sin x \\cos x) - (2 \\sin x \\cos x)^2 \\sin x \\, dx \\\\ &= \\int 2 \\cos^4 x \\sin x - 2 \\cos^2 x \\sin^3 x - 4 \\sin^3 x \\cos^2 x \\, dx \\\\ &= \\int 2 \\cos^4 x \\sin x - 6 \\sin^3 x \\cos^2 x \\, dx \\\\ &= \\int 2 \\cos^4 x \\sin x \\, dx - \\int ... "}, {"docid": 219019, "score": 9.151, "title": "Integration of trig function", "url": "http://math.stackexchange.com/questions/2458459/integration-of-trig-function", "snippet": "then into [imath](\\cos x-\\sin x)*(\\cos x+\\sin x)[\/imath]. It becomes [imath](\\cos x-\\sin x)^{11}(\\cos x+\\sin x)[\/imath] From here I am stuck in substitution. Thanks in advance No, it becomes [imath](\\cos x-\\sin x)^{11}(\\cos x+\\sin x)^{10}[\/imath]. Interesting problem. You almost did it. So we have \\begin{align} I&=\\int<\/em>(\\cos<\/em> x<\/em>-\\sin<\/em> x<\/em>)\\cos<\/em>^{10}2x<\/em>\\,dx=\\int<\/em>(\\cos<\/em> x<\/em>-\\sin<\/em> x<\/em>)(\\cos<\/em>^2x<\/em>-\\sin<\/em>^2x<\/em>)^{10}dx\\\\ &=\\int<\/em>(\\cos<\/em> x<\/em>-\\sin<\/em> x)^{11}(\\cos<\/em> x+\\sin<\/em> x)^{10}dx. \\end{align} Now notice \\begin{align} \\cos<\/em> x+\\sin<\/em> x&=\\sqrt{2}\\sin(x+\\frac{\\pi}{4}),\\\\ \\cos x-\\sin x&=\\sqrt{2}\\cos(x+\\frac{\\pi}{4}). \\end{align} Therefore, we have \\begin{align} I&=2^{21\/2}\\int\\cos^{11}(x+\\frac{\\pi}{4})\\sin^{10}(x+\\frac{\\pi}{4})dx\\\\ &=1024\\sqrt{2}\\int\\cos^{10}(x+\\frac{\\pi}{4})\\sin^{10}(x+\\frac{\\pi}{4})\\,d\\sin(x+\\frac{\\pi}{4})\\\\ &=1024\\sqrt{2}\\int\\lef ... "}, {"docid": 25364, "score": 9.085, "title": "what's the approach used in this integral?", "url": "http://math.stackexchange.com/questions/2286759/whats-the-approach-used-in-this-integral", "snippet": "what's the approach used in this integral? in : [imath]\\displaystyle \\int (\\sin 3x)(\\cos 5x)\\, dx [\/imath] ( which recurred as hyperbolic sin<\/em> and cosine as well ) consider [imath]\\sin(ax+bx)[\/imath] then consider [imath]\\sin(ax-bx)[\/imath] Can you combine them in some way to simplify your problem. Apparently, [imath]\\cos(s)*\\sin(t) = \\frac{\\sin(s+t) - \\sin(s-t)}{2}[\/imath], which should help. See mathworld.wolfram.com\/WernerFormulas.html You can use t ... cos 5x[\/imath] This means that [imath]2\\sin3x \\cos 5x = \\sin 8x - \\sin 2x[\/imath], and from here you should be able to finish off the answer on your own. Someone already suggested using the identity [imath]\\sin A \\cos B = \\frac{ \\sin(A-B) + \\sin(A+B)} 2.[\/imath] That's one way. Another way is this: \\begin{align} & \\int<\/em> \\sin<\/em>(3x<\/em>)\\cos<\/em>(5x<\/em>)\\,dx = \\int<\/em> \\frac<\/em>{e^{3ix} - e^{-3ix}}{2i} \\cdot \\frac<\/em>{e^{5ix}+ e^{-5ix}} 2 \\,dx \\\\[10pt] = {} & \\frac<\/em> 1 {4i} \\int<\/em> (e^{8ix} - e^{2ix} - e^{-2ix} + e^{-8ix}) \\,dx \\end{align} then integrate term by term and then use [imath]e^{i\\theta} = \\cos\\theta + i\\sin\\theta[\/imath] to get a function of sines and cosines. And here's another way: \\begin{align} & \\int<\/em> \\sin<\/em>(3x<\/em>)\\Big(\\cos<\/em>(5x<\/em>)\\,dx\\Big) = \\overbrace{\\int<\/em> u\\,dv = uv - \\int<\/em> v\\,du}^\\text{integration by parts} \\\\[10pt] = {} & \\sin<\/em>(3x<\/em>) \\frac<\/em>{\\sin<\/em>(5x<\/em>)} 5 - \\int<\/em> \\frac {\\sin(5x)} 5 \\cdot 3\\cos(3x)\\,dx \\\\[10pt] = {} & \\frac 1 5 \\sin(3x)\\sin(5x) - \\frac 3 5 \\int \\cos(3x) \\Big(\\sin(5x)\\,dx\\Big) \\\\[10pt] = {} & \\frac 1 5 \\sin(3x)\\sin(5x) - \\frac 3 5 \\left( \\int s\\,dt \\right) \\\\[10pt] = {} & \\frac 1 5 \\sin(3x)\\sin(5x) - \\frac 3 5 \\left( st - \\int t \\,ds \\right) \\\\[10pt] = {} & \\fr ... "}]}