Question

If $\int \left\{\sin \right(101x)\cdot {\sin }^{99}x\}dx=\frac{\sin \left(100x\right)(\sin x{)}^{25\lambda }}{10\mu }+C,$ then the value of $(\lambda +\mu )$ equals to (where $\lambda ,\mu$ are fixed constants and $C$ is constant of integration)

Answer 1

Given : $\int \left\{\sin \right(101x)\cdot {\sin }^{99}x\}dx=\frac{\sin \left(100x\right)(\sin x{)}^{25\lambda }}{10\mu }+C\cdots \left(i\right)$
Now, $L.H.S.=\int \left\{\sin \right(101x)\cdot {\sin }^{99}x\}dx=\int \left\{\sin \right(100x+x)\cdot {\sin }^{99}x\}dx=\int \left\{\sin \right(100x)\cdot \cos x+\cos (100x\left)\sin x\right\}(\sin x{)}^{99}dx=\int \sin \left(100x\right)\cdot \cos x\cdot {\sin }^{99}xdx+\int \cos \left(100x\right){\sin }^{100}xdx=I_1+I_2\cdots \left(ii\right)$
In first integral, applying integration by part , by taking $u=\sin \left(100x\right),v=\cos x\cdot {\sin }^{99}x$
As,$\int v\cdot dx=\int {\sin }^{99}x\cdot d\left(\sin x\right)=\frac{{\sin }^{100}x}{100}$
$\Rightarrow I_1=\frac{\sin \left(100x\right)\cdot {\sin }^{100}x}{100}-\int \cos \left(100x\right){\sin }^{100}xdx$
$\Rightarrow I_1=\frac{\sin \left(100x\right)\cdot {\sin }^{100}x}{100}-I_2$
Substituting in $\left(ii\right):$
$\Rightarrow L.H.S=\frac{\sin \left(100x\right)\cdot {\sin }^{100}x}{100}$
Comparing with $R.H.S.$ of $\left(i\right):$
We get, $\lambda =4,\mu =10$
$\therefore \lambda +\mu =14$