Question

$\int_{0}^{1}\left(\frac{x^{15}}{1+x^{32}}\right)[\cos(\tan^{-1}x^{16})]dx$ is equal to

• A. $\frac{1}{2\sqrt{2}}$
• B. $\frac{1}{8\sqrt{2}}$
• C. $\frac{1}{32\sqrt{2}}$
• D. $\frac{1}{4\sqrt{2}}$
• E. $\frac{1}{16\sqrt{2}}$

Hint:

Math Tip: Whenever you see an inverse trigonometric function as an argument, look for its derivative elsewhere in the integrand. Here, substituting $\theta = \tan^{-1}(x^{16})$ handles both the complex argument and the rational fraction in one unified step!

Answer 1

Answer: (E)

Concept:
Calculus - Definite Integration by Substitution.
Step 1: Identify the substitution variable.
Let $t = x^{16}$.
Differentiate both sides with respect to $x$: $$ dt = 16x^{15} dx $$ $$ \frac{dt}{16} = x^{15} dx $$
Step 2: Change the limits of integration.

• When $x = 0$, $t = (0)^{16} = 0$.
• When $x = 1$, $t = (1)^{16} = 1$.

The new limits for $t$ are from $0$ to $1$.
Step 3: Substitute into the integral.
Replace $x^{16}$ with $t$, $x^{32}$ with $t^2$, and $x^{15}dx$ with $\frac{dt}{16}$: $$ I = \int_{0}^{1} \frac{1}{1+t^2} \cos(\tan^{-1}t) \frac{dt}{16} $$ $$ I = \frac{1}{16} \int_{0}^{1} \cos(\tan^{-1}t) \frac{1}{1+t^2} dt $$
Step 4: Perform a second substitution.
Let $\theta = \tan^{-1}t$.
Differentiate with respect to $t$: $$ d\theta = \frac{1}{1+t^2} dt $$
Step 5: Change the limits for the second substitution.

• When $t = 0$, $\theta = \tan^{-1}(0) = 0$.
• When $t = 1$, $\theta = \tan^{-1}(1) = \frac{\pi}{4}$.

Step 6: Evaluate the final integral.
Substitute $\theta$ and $d\theta$ into the integral: $$ I = \frac{1}{16} \int_{0}^{\pi/4} \cos\theta \,d\theta $$ Integrate $\cos\theta$: $$ I = \frac{1}{16} [\sin\theta]_{0}^{\pi/4} $$ Substitute the upper and lower limits: $$ I = \frac{1}{16} \left( \sin\left(\frac{\pi}{4}\right) - \sin(0) \right) $$ $$ I = \frac{1}{16} \left( \frac{1}{\sqrt{2}} - 0 \right) = \frac{1}{16\sqrt{2}} $$