Indefinite Integration
Substitution + Boundary Condition
nta_pyq_2024_jan
Grade 12
Question:
For $x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, if $y(x)=\displaystyle\int\dfrac{\csc x+\sin x}{\csc x\sec x+\tan x\sin^2x}\,dx$ and $\lim_{x\to(\pi/2)^-}y(x)=0$, then $y\left(\dfrac{\pi}{4}\right)$ is equal to
$\tan^{-1}\!\left(\dfrac{1}{\sqrt{2}}\right)$
$\dfrac{1}{2}\tan^{-1}\!\left(\dfrac{1}{\sqrt{2}}\right)$
$-\dfrac{1}{\sqrt{2}}\tan^{-1}\!\left(\dfrac{1}{\sqrt{2}}\right)$
$\dfrac{1}{\sqrt{2}}\tan^{-1}\!\left(-\dfrac{1}{\sqrt{2}}\right)$
Step-by-Step Solution
Key Concept: Simplify the integrand: $\frac{\csc x+\sin x}{\csc x\sec x+\tan x\sin^2x}=\frac{(1+\sin^2x)\cos x}{1+\sin^4x}$. Let $t=\sin x$, reduce to $\int\frac{1+t^2}{1+t^4}dt$ and use the standard form $\int\frac{1+1/t^2}{(t-1/t)^2+2}dt$.
Simplify: integrand $=\frac{(1+\sin^2x)\cos x}{1+\sin^4x}$. Sub $t=\sin x$: $\frac{1}{\sqrt2}\tan^{-1}\!\left(\frac{t-1/t}{\sqrt2}\right)$. BC: $C=0$. $y(\pi/4)=-\frac{1}{\sqrt2}\tan^{-1}\!\left(\frac{1}{\sqrt2}\right)$.
Correct Answer: 3