Integration of Sinx

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 ?

What is the Integration of Sin Inverse x ?

What is the Differentiation of sin inverse x ?

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