Question

Solve :

$\int \left(\sin \right(101x).{\sin }^{99}x)dx$

• A. $\frac{\sin \left(100x\right)(\sin x{)}^{100}}{100}+C$
• B. $\frac{\cos \left(100x\right)(\sin x{)}^{100}}{100}+C$
• C. $\frac{\cos \left(100x\right)(\cos x{)}^{100}}{100}+C$
• D. $\frac{\sin \left(100x\right)(\sin x{)}^{101}}{101}+C$

Answer 1

The correct option is A $\frac{\sin \left(100x\right)(\sin x{)}^{100}}{100}+C$

$I=\int \left(\sin \right(101x).{\sin }^{99}x)dx=\int \left(\sin \right(100x+x)\cdot {\sin }^{99}x)$

$=\int (\sin 100x\cos x+\cos 100x\sin x).{\sin }^{99}xdx$ $[\because \sin (A+B)=\sin A\cos B+\sin B\cos A]$

Let $t=\sin 100x{\sin }^{100}x$

$\Rightarrow dt=\sin 100x\times 100\times {\sin }^{99}x\cos x+{\sin }^{100}x\cos 100x\times 100$

$dt=100(\sin 100x\cos x+\cos 100x\sin x){\sin }^{99}xdx$

Now, $I=\int \frac{dt}{100}=\frac{t}{100}+C=\frac{\sin 100x{\sin }^{100}x}{100}+C$
Ans: A