Question

evaluate the integrals?
chapter - integration ​

Answer 1

Topic

Integration

Question :-

$\int \texttt{sin(mx).sin(nx) dx}$

Solving

We can write integrand as

$\frac{1}{2}\int \texttt{cos(m - n)x - cos(m + n)x dx}$

from Trigonometric Formula,

2sinA.sinB = cos( A - B ) - cos( A + B )

Now, we can write

$\frac{1}{2}\int \texttt{cos(m - n)x - cos(m + n)x dx}$

as

$\frac{1}{2}\int\texttt{(u + v) dx}$

where

$\texttt{u = cos(m - n)x and}$

$\texttt{v = cos(m + n)x}$

We know that

$\int\texttt{coskx dx} = \frac{1}{\texttt{k}}\texttt{sinkx +  C}$

So,

$\int\texttt{u dx} = \int \texttt{cos(m - n)x} = \frac{1}{\texttt{(m - n)}}\texttt{sin(m - n)x +  C}$

$\int\texttt{v dx} = \int \texttt{cos(m + n)x} = \frac{1}{\texttt{(m + n)}}\texttt{sin(m + n)x +  C}$

Answer

$\int \frac{1}{2}\texttt{(u + v) dx} = \frac{1}{2}\int\texttt{u dx} + \frac{1}{2}\int \texttt{v dx}$

$\frac{1}{\texttt{2(m - n)}}\texttt{sin(m - n)x} + \frac{1}{\texttt{2(m + n)}}\texttt{sin(m + n)x + C'}$