# Integral of trigonometric function.

The question is:

$$\int^{\pi/4}_{0}\left(\cos 2\theta \right) ^{3/2}\cos \theta\, d\theta$$

I tried to write it in terms of $\sin\theta$ and the substitute $\sin\theta=t$ but then got stuck at the subsequent integral. I tried integrating by parts, which was of no use.

Any help would be appreciated. Thanks.

## 2 Answers

we will make a change of variable $x = \sqrt 2\sin \theta, dx = \sqrt 2 \cos \theta \, d\theta$ with this we have \begin{align} \int^{\pi/4}_{0}\left(\cos 2\theta \right) ^{\frac{3}{2}}\cos \theta d\theta &= \frac 1 {\sqrt 2}\int_0^1 (1 - x^2)^{3/2} \, dx \\ &= \frac 1{\sqrt 2}\left((1-x^2)^{3/2}x\big|_0^1 + 3\int_0^1x^2(1-x^2)^{1/2}\right)\, dx\\ &= \frac 3{\sqrt 2}\int_0^1x^2(1-x^2)^{1/2}\, dx, x = \sin u\\ &=\frac 3{\sqrt 2}\int_0^{\pi/2}\sin ^2u\cos^2u\, du\\ &=\frac 3{4\sqrt 2}\int_0^{\pi/2}\sin ^2 2u\, du = \frac 3{8\sqrt 2}\int_0^{\pi/2}(1 - \cos 4u)\, du\\ &= \frac {3\pi}{16\sqrt 2} \end{align}

I propose $sin(\theta)=\dfrac{u}{\sqrt{2}}$, then $u=sin(t)$.