Question

Given $\int \sin \left(x\right)dx=-\cos \left(x\right)$ and $\int \cos \left(x\right)dx=\sin \left(x\right)$. If $f\left(x\right)=36\int \left(\sin \left(2x\right)+\cos \left(3x\right)\right]dx$, then find the value of $-f\left(\pi \right)$ [ take constant of integration equal to zero]

___

Answer 1

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

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

We are given $\int \sin \left(x\right)dx=-\cos \left(x\right)$ and $\int \cos \left(x\right)dx=\sin \left(x\right)$

We want to find, $-f\left(\pi \right)$ For this we will find f(x) first

We have,

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

Let’s consider $\int \left(\sin \left(2x\right)+\cos \left(3x\right)\right]dx$

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

We will now find these two integrals separately and proceed

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

= 12 sin3x -18cos2x.

We are not including the constant of integration, because it is given as zero in question.

So we get f(x) = 12 sin3x -18cos2x

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