MHB How Do You Evaluate the Integral of Sin^4(x) from 0 to Pi?

[karush]

206.8.3.9
$\displaystyle
I_9=\int_{0}^{\pi} \sin^4\left({\theta}\right) \,d\theta
=\frac{3\pi}{8}$
$$I_9
=\int_{0}^{\pi} \frac{(1-\cos\left({2\theta}\right))^2}{4}\,d\theta $$
Expand
$$\displaystyle
I_9
=\frac{1}{4}\left[
\int 1 \,d\theta
-2\int \cos\left({2\theta}\right) \,d\theta
+\int \cos^2\left({2\theta}\right) \,d\theta
\right]_0^\pi$$
integrat
$$\displaystyle
I_9
=\frac{1}{4}\left[
\theta
-\frac{\sin\left({2\theta}\right)}{2}
+\frac{\sin\left({4\theta}\right)}{8}
+\frac{\theta}{2}
\right]_0^\pi=\frac{3\pi}{8}$$

 

[kaliprasad]

206.8.3.9
$\displaystyle
I_9=\int_{0}^{\pi} \sin^4\left({\theta}\right) \,d\theta
=\frac{3\pi}{8}$
$$I_9
=\int_{0}^{\pi} \frac{(1-\cos\left({2\theta}\right))^2}{4}\,d\theta $$
Expand
$$\displaystyle
I_9
=\frac{1}{4}\left[
\int 1 \,d\theta
-2\int \cos\left({2\theta}\right) \,d\theta
+\int \cos^2\left({2\theta}\right) \,d\theta
\right]_0^\pi$$
integrat
$$\displaystyle
I_9
=\frac{1}{4}\left[
\theta
-\frac{\sin\left({2\theta}\right)}{2}
+\frac{\sin\left({4\theta}\right)}{8}
+\frac{\theta}{2}
\right]_0^\pi=\frac{3\pi}{8}$$

Above approach is right. what do you want to know ?

 

[karush]

I saw some other solutions to this but they used power reduction it really got confusing.