Question

If the integral of the function $\sin \left(80x\right){\sin }^{78}x$ is $\frac{\sin ax{\sin }^{b}x}{c}+D,$ where $a,b,c\in R$ and $D$ is the constant of integration, then the value of $a+b+c$ is equal to

• A. $79$
• B. $81$
• C. $237$
• D. $240$

Answer 1

The correct option is C $237$

$I=\int \sin \left(80x\right){\sin }^{78}xdx$
$=\int \sin (79x+x){\sin }^{78}xdx$
$=\int \left(\sin \right(79x)\cos x+\sin (x\left)\cos 79x\right){\sin }^{78}xdx$
$=\int \left(\sin \right(79x){\sin }^{78}x\cos x+{\sin }^{79}x\cos 79x)dx$

Integrating by parts, we get
$I=\frac{\sin \left(79x\right){\sin }^{79}x}{79}-\frac{79}{79}\int \cos \left(79x\right){\sin }^{79}xdx+\int \cos \left(79x\right){\sin }^{79}xdx$
$=\frac{\sin \left(79x\right){\sin }^{79}x}{79}+D$
$\therefore a+b+c=237$