This question already has an answer here:

I have the basic idea of how to work out the integral of a trig function, but am having trouble in applying the concept. Would really appreciate it if someone could help me. Thanks!


marked as duplicate by Hans Lundmark, Tom-Tom, Claude Leibovici calculus Jan 27 '16 at 8:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


$\int 4 \cos^2(x) dx$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=4\int\cos^2(x) dx$

$\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)=\frac{1+\cos \left(2x\right)}{2}$

$=4\int \frac{1+\cos \left(2x\right)}{2}dx$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=4\frac{1}{2}\int \:1+\cos \left(2x\right)dx$

$\mathrm{Apply\:Integral\:Substitution:}\:\int > f\left(g\left(x\right)\right)\cdot g^{'}\left(x\right)dx=\int > f\left(u\right)du,\:\quad u=g\left(x\right)$



$\quad \Rightarrow \:du=2dx$

$\Rightarrow \:dx=\frac{1}{2}du$

$=\int \left(1+\cos \left(u\right)\right)\frac{1}{2}du$

$=4\frac{1}{2}\int \left(1+\cos \left(u\right)\right)\frac{1}{2}du$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=4\frac{1}{2}\frac{1}{2}\int \:1+\cos \left(u\right)du$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=4\frac{1}{2}\frac{1}{2}\left(\int \:1du+\int \cos \left(u\right)du\right)$

$=4\frac{1}{2}\frac{1}{2}\left(u+\sin \left(u\right)\right)$


$=4\frac{1}{2}\frac{1}{2}\left(2x+\sin \left(2x\right)\right)$


$=2x+\sin \left(2x\right)$


$=2x+\sin \left(2x\right)+C$

  • 1
    $\begingroup$ Mithilesh, I can't thank you enough for such answer. I hope you have a good day. Your answer helped me a lot more than you'd think. Thanks again. $\endgroup$ – Krunal Rindani Jan 27 '16 at 11:02
  • $\begingroup$ @KrunalRindani, you are welcome :). $\endgroup$ – Mithlesh Upadhyay Jan 27 '16 at 11:49

Not the answer you're looking for? Browse other questions tagged or ask your own question.