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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  • $\begingroup$ thank you so much! $\endgroup$ – hahaha Feb 12 '19 at 20:44
  • $\begingroup$ Wolfram gives the value $\frac{1}{20}\left(3\pi - \ln(9)\right)$ $\endgroup$ – aleden Feb 12 '19 at 20:50
  • $\begingroup$ Hint: Divide by $\cos t$ then let $\tan t =x$ and use partial fractions. $\endgroup$ – Zacky Feb 12 '19 at 21:34

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 '19 at 20:59
  • $\begingroup$ i tried to times numerator and denominator by sec^3(t) $\endgroup$ – hahaha Feb 12 '19 at 21:00
  • $\begingroup$ Do you see it now? $\endgroup$ – 高田航 Feb 12 '19 at 21:08
  • $\begingroup$ substitute in into function ? $\endgroup$ – hahaha Feb 12 '19 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 '19 at 22:09

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