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Compute the integral, I cannot use $u$-substitution to compute this. $$\int_0^{\pi/2}\frac{\cos(t)}{3\cos(t)+\sin(t)}dt$$

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closed as off-topic by Lee David Chung Lin, Gibbs, trancelocation, Aweygan, Song Feb 13 at 20:19

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    $\begingroup$ Welcome to MSE. Please type your questions when possible, instead of posting links. Links can't be browsed, and can get broken. also, you'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments. $\endgroup$ – saulspatz Feb 12 at 20:41
  • $\begingroup$ thank you so much! $\endgroup$ – hahaha Feb 12 at 20:44
  • $\begingroup$ Wolfram gives the value $\frac{1}{20}\left(3\pi - \ln(9)\right)$ $\endgroup$ – aleden Feb 12 at 20:50
  • $\begingroup$ Hint: Divide by $\cos t$ then let $\tan t =x$ and use partial fractions. $\endgroup$ – Number Feb 12 at 21:34
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A good idea when tackling these sine and cosine problems is to divide by $\cos x$ to create $\tan x$ in the integrand and let $t=\tan x$. Indeed, let's give that a try. If we divide by $\cos x$ in the integrand, then$$\frac {\cos x}{3\cos x+\sin x}=\frac 1{3+\tan x}$$Now let $t=\tan x$ and observe that $x=\arctan t$. Therefore$$\mathrm dx=\frac {\mathrm dt}{1+t^2}$$

Thus$$\begin{align*}\int\mathrm dx\,\frac {\cos x}{3\cos x+\sin x} & =\int\frac {\mathrm dx}{3+\tan x}\\ & =\int\frac {\mathrm dt}{(1+t^2)(3+t)}\end{align*}$$It's easy to verify that the integrand is also equal as$$\frac 1{(1+t^2)(3+t)}=\frac 1{10(3+t)}-\frac {t}{10(1+t^2)}+\frac 3{10(1+t^2)}$$Integrating each term, then$$\begin{align*}I & =\frac 1{10}\log(t+3)-\frac 1{20}\log(1+t^2)+\frac 3{10}\arctan t+C\end{align*}$$Substitute back $t$ with $\tan x$ to complete the proof.

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Hint: $$\cos(x)=\frac{3}{10}(\sin(x)+3\cos(x))+\frac{1}{10}(\cos(x)-3\sin(x))$$

This is because \begin{align*} \cos(x)& =\frac{3}{10}(\sin(x)+3\cos(x))+\frac{1}{10}(\cos(x)-3\sin(x)) \\ &= \frac{3}{10}\sin(x)-\frac{3}{10}\sin(x)+\frac{9}{10}\cos(x)+\frac{1}{10}\cos(x) \\ & = \cos(x) \end{align*}

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  • $\begingroup$ i don't quite understand that. $\endgroup$ – hahaha Feb 12 at 20:59
  • $\begingroup$ i tried to times numerator and denominator by sec^3(t) $\endgroup$ – hahaha Feb 12 at 21:00
  • $\begingroup$ Do you see it now? $\endgroup$ – 高田航 Feb 12 at 21:08
  • $\begingroup$ substitute in into function ? $\endgroup$ – hahaha Feb 12 at 21:15
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    $\begingroup$ You forgot to put a minus in the second bracket. Equality currently doesn't hold! $\endgroup$ – Peter Foreman Feb 12 at 22:09

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