Question

$f\left(x\right)=36\int \left[\sin \right(2x)+\cos (3x\left)\right]dx$, then the value of $f\left(\pi \right)$= if the constant of integration is equal to zero.

• A. 18
• B. 36
• C. -18
• D. -36

Answer 1

The correct option is A 18

We knowthat $\int \left(f\right(x)+g(x\left)\right)dx=\int f\left(x\right)dx+\int g\left(x\right)dx$ and $\int f(ax+b)dx=\frac{F(ax+b)}{a}+c$, where $\int f\left(x\right)dx=F\left(x\right)$

We will solve this problem using these two theorems on integration.

We know that $\int \sin xdx=-\cos x$ and $\int \cos xdx=\sin x.$

We have,

f(x) = 36 $\int \left[\sin \right(2x)+\cos (3x\left)\right]dx$

$=\int \sin \left(2x\right)dx+\int \cos \left(3x\right)dx$

$\int \sin \left(2x\right)dx=\frac{-\cos 2x}{2}$ and $\int \cos \left(3x\right)dx=\frac{\sin 3x}{3}$

$\Rightarrow 36\int \left[\sin \right(2x)+\cos (3x\left)\right]dx=36\frac{\sin 3x}{3}+36\frac{-\cos 2x}{2}$

$=12\sin \left(3x\right)-18\cos \left(2x\right)+c$.

$=12\sin \left(3x\right)-18\cos \left(2x\right)$ [Since c = 0]

$\Rightarrow f\left(\pi \right)=12\sin \left(3\pi \right)-18\cos \left(2\pi \right)=12\times 0-18\times 1=-18$