Here you will learn proof of integration of sinx or sin x and examples based on it.
Let’s begin –
Integration of Sinx or Sin x
The integration of sinx is -cosx + C
where C is the integration constant.
i.e. \(\int\) (sinx) dx = -cos x + C
Proof :
We will prove this formula using differentiation,
Let \(d\over dx\)(-cos x + C) = -\(d\over dx\) cos x + \(d\over dx\) C
Using differentiation formula,
\(d\over dx\) cos x = -sin x and differentation of constant is 0.
\(\implies\) \(d\over dx\)(-cos x + C) = -\(d\over dx\) cos x + \(d\over dx\) C
\(\implies\) \(d\over dx\)(-cos x + C) = -(-sin x) + 0
We can also write it as,
sin x = \(d\over dx\)(-cos x + C)
Now, integrating on both sides,
\(\int\) sin x = \(\int\) \(d\over dx\)(-cos x + C)
We know that integration and differentiation both are reciprocals of each other, so in right hand side expression they cancel each other and we get,
Hence, \(\int\) sin x = -cos x + C
Example : Prove that \(\int\) sin (ax + b) = \(-1\over a\) cos(ax + b) + C.
Solution : We have,
I = \(\int\) sin (ax + b) dx
Let ax + b = t, Then , d(ax + b) = dt \(\implies\) adx = dt
\(\implies\) dx = \(1\over a\) dt
Putting ax + b = t and dx = \(1\over a\) dt , we get
\(\int\) sin (ax + b) dx = \(1\over a\) \(\int\) sin t dt
= \(-1\over a\) cos t + C
= \(-1\over a\) cos(ax + b) + C
Hence, \(\int\) sin (ax + b) = \(-1\over a\) cos(ax + b) + C
Related Questions
What is the Differentiation of sin x ?
