JEE Advanced · Mathematics · 26. Indefinite Integration
Let \(F(x)\) be an indefinite integral of \(\sin ^2 x\).
Statement I The function \(F(x)\) satisfies \(F(x+\pi)=F(x)\) for all real \(x\).
Statement II \(\sin ^2(x+\pi)=\sin ^2 x\) for all real \(x\).
- A Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I
- B Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I.
- C Statement I is true, Statement II is false
- D Statement I is false, Statement II is true
Answer & Solution
Correct Answer
(D) Statement I is false, Statement II is true
Step-by-step Solution
Detailed explanation
\[
\begin{aligned}
& F(x)=\int \sin ^2 x d x=\int \frac{1-\cos 2 x}{2} d x \\
& \Rightarrow F(x)=\frac{1}{4}(2 x-\sin 2 x)+c
\end{aligned}
\]
Since, \(F(x+\pi) \neq F(x)\).
Hence, Statement I is false.
But Statement II is true as \(\sin ^2 x\) is periodic with period \(\pi\).
\begin{aligned}
& F(x)=\int \sin ^2 x d x=\int \frac{1-\cos 2 x}{2} d x \\
& \Rightarrow F(x)=\frac{1}{4}(2 x-\sin 2 x)+c
\end{aligned}
\]
Since, \(F(x+\pi) \neq F(x)\).
Hence, Statement I is false.
But Statement II is true as \(\sin ^2 x\) is periodic with period \(\pi\).
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