Question

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

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Answer 1

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

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

We are given $\int$ sinx dx = -cosx and $\int$ cosx dx = sinx.

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

We have,

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

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

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

We will now find these two integrals separately and proceed

$\int$ sin (2x) $=\frac{-cos2x}{2}$ and $\int$cos (3x) = $\frac{sin3x}{3}$

$\Rightarrow$ 36 $\int$ [sin (2x) + cos (3x)] = $36\frac{sin3x}{3}+36\frac{-cos2x}{2}$

= 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

$\Rightarrow f\left(\pi \right)=12sin3\left(\pi \right)-18cos2\left(\pi \right)=12\times 0-18\times 1=-18\Rightarrow -f\left(\pi \right)=18$