Question on Integration by parts

Question

4. Evaluate the integral $$\int_{-1}^{1}\frac{1}{(x^2+1)^3}\,dx$$You will have to apply integration by parts in some of the intermediate steps.
5. b) Use the $t$-substitution $t = \tan\frac\theta2$, $\cos\theta = \frac{1-t^2}{1+t^2}$, $\sin\theta = \frac{2t}{1+t^2}$, $\frac{d\theta}{dt} = \frac{2}{1+t^2}$ to evaluate $$\int_{0}^{\frac{\pi}{2}} \frac{1}{2+\cos x}\,d\theta$$

For Question $4$: I substitute $x=\tan \theta$ and the fact that $dx=\sec^2 \theta d\theta$

After simplifying, I was asked to integrate $(\cos \theta)^4 d\theta$ which doesn't seem right.

The answer for Question $4$ is $\frac12 + \frac{3\pi}{16}$

For Question $5(b)$: I substitute in the values and got $\frac2{3+t^2}$, then I integrated that and solve which didn't lead me to the right answer $\frac{\pi}{3\sqrt{3}}$

Any help would be greatly appreciated. Thank you so much.

Answer 1

For Question 4: I substitute x=tanθ and the fact that dx=sec^2θ dθ

After simplifying, i was asked to integrate (cosθ)^4 dθ which doesn't seem right.

Nope. It is okay. $$\begin{align}\int_{-1}^1 (x^2+1)^{-3}\operatorname d x ~=~ & 2\int_0^1(x^2+1)^{-3}\operatorname d x & \\[1ex] ~=~ 2 &\int_{0}^{\pi/4} \big(\tan^2(\theta)+1\big)^{-3}\cdot \sec^2 (\theta) \operatorname d \theta \\[1ex] ~=~& 2\int_{0}^{\pi/4} \big(\sec^2(\theta)\big)^{-3}\cdot \sec^2(\theta)\operatorname d \theta \\[1ex] ~=~& 2\int_{0}^{\pi/4} \cos^4 \theta \operatorname d \theta\end{align}$$

Now use your trig identities. $$\cos^2 \theta = \tfrac 12(1+\cos 2\theta)$$

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Alternatively use $\cos\theta = \tfrac 12(\mathsf e^{\imath\theta}+\mathsf e^{-\imath\theta})$

Answer 2

Hint.

• To evaluate $\displaystyle \int_{-1}^1\frac1{(x^2+1)^3}\:dx$ the change of variable $x=\tan \theta$ gives

  $$ \begin{align} \int_{-1}^1\frac1{(x^2+1)^3}\:dx&=2\int_0^1\frac1{(x^2+1)^3}\:dx \\\\&=2\int_0^{\pi/4}\frac{(\tan^2\theta+1)\:d\theta}{(\tan^2\theta+1)^3} \\\\&=2\int_0^{\pi/4}\cos^4\!\theta \:d\theta \\\\&=\frac12\int_0^{\pi/4}(1+\cos (2\theta))^2 \:d\theta \\\\&=\frac12\int_0^{\pi/4}(1+2\cos (2\theta)+\cos^2(2\theta)) \:d\theta \end{align} $$ then it is easier to obtain the evaluation.

• To evaluate $\displaystyle \int_0^{\pi/2}\frac1{2+\cos \theta}\:d\theta$ the change of variable $t=\tan \frac{\theta}2$ gives

  $$ \begin{align} \int_0^{\pi/2}\frac1{2+\cos \theta}\:d\theta&=2\int_0^1\frac1{3+t^2}\:dt \\\\&=2\left[\frac{1}{\sqrt{3}}\arctan \frac{t}{\sqrt{3}} \right]_0^1 \end{align} $$ then it is easier to obtain the evaluation.