# integral with trig substitution

Problem:

Integrate : $$\displaystyle \int^\pi_0\hspace{1mm}\sin^2t\cos^4t~dt$$

What I tried :

1. Use the pythagorean identity $$\cos^2t=1-\sin^2t$$ to rewrite the problem as $$\int^\pi_0\hspace{1mm}\sin^2t\hspace{1mm}(1-\sin^2t)\cos^2t\,dt$$

2. Distribute $$\sin^2t$$ to rewrite the problem as

$$\int^\pi_0\hspace{1mm}(\sin^2t-\sin^4t)\cos^2t \, dt$$

3. let $$u = \sin^2t$$ ; $$du=2\cos^2t\hspace{1mm}dt$$

4. Make the u substitution to rewrite the problem as

$$\frac 1 2 \int^{\sin^2\pi}_{\sin^20}\hspace{1mm}u - u^2\,du$$

When evaluated the whole thing obviously just goes to zero. When I look in the back of the book the correct answer is $$\hspace{2mm}\pi/16$$

• You substituted $u = \sin^2 t$. So $du = 2 \sin t \cos t \ dt$. Commented Sep 21, 2017 at 3:19

Hint:) Use identities $$\sin^2t=\dfrac{1-\cos2t}{2}~~~~,~~~~\cos^2t=\dfrac{1+\cos2t}{2}$$ and reduce powers.
• You made a mistake in $du$ as @samjoe said in his comments, anyway this substitution isn't applicable since it need a term of $cos$ with power $1$ only. Commented Sep 21, 2017 at 3:41