Question

Assertion :Let F(x) be an indefinite integral of $si{n}^{8}x$.
Statement 1: $F\left(x\right)-\frac{35}{128}x$ is a periodic function with period $\pi$. Reason: Statement 2: The period of $si{n}^n\alpha x$ is $2\pi /a$.

• A. Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation for Statement 1
• B. Both Statement 1 and Statement 2 are correct but Statement 2 is not the correct explanation for Statement 1
• C. Statement 1 is correct but Statement 2 is incorrect
• D. Both Statement 1 and Statement 2 are incorrect

Answer 1

The correct option is A Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation for Statement 1

Let $I=\int {\sin }^{8}xdx$
Using $\int {\sin }^{m}dx=\frac{-\cos x{\sin }^{m-1}x}{m}+\frac{m-1}{m}\int {\sin }^{m-2}xdx$
$I=-\frac{1}{8}{\sin }^{7}x\cos x+\frac{7}{8}\int {\sin }^{6}xdx$
$=-\frac{1}{8}{\sin }^{7}x\cos x-\frac{7}{48}{\sin }^{5}x\cos x+\frac{35}{48}\int {\sin }^{4}xdx$
$=-\frac{1}{8}{\sin }^{7}x\cos x-\frac{7}{48}{\sin }^{5}x\cos x-\frac{35}{192}{\sin }^{3}x\cos x$
$+\frac{35}{64}\int {\sin }^{2}xdx$
$=-\frac{1}{8}{\sin }^{7}x\cos x-\frac{7}{48}{\sin }^{5}x\cos x-\frac{35}{192}{\sin }^{3}x\cos x$
$+\frac{35}{64}\int (\frac{1}{2}-\frac{1}{2}\cos 2x)dx$
$=-\frac{1}{8}{\sin }^{7}x\cos x-\frac{7}{48}{\sin }^{5}x\cos x-\frac{35}{192}{\sin }^{3}x\cos x$
$-\frac{35}{256}\sin 2x+\frac{35x}{128}$