Integration of Cosx

Here you will learn proof of integration of cosx or cos x and examples based on it.

Let’s begin –

Integration of Cosx or Cos x

The integration of cosx is sinx + C

where C is the integration constant.

i.e. \(\int\) (cosx) dx = sin x + C

Proof :

We will prove this formula using differentiation,

Let \(d\over dx\)(sin x + C) = \(d\over dx\) sin x + \(d\over dx\) C

Using differentiation formula,

\(d\over dx\) sin x = cos x and differentation of constant is 0.

\(\implies\) \(d\over dx\)(sin x + C) = \(d\over dx\) sin x + \(d\over dx\) C

\(\implies\) \(d\over dx\)(sin x + C) = cos x + 0

We can also write it as,

cos x = \(d\over dx\)(sin x + C)

Now, integrating on both sides,

\(\int\) cos x = \(\int\) \(d\over dx\)(sin 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, \(\int\) cos x = sin x + C

Example : Prove that \(\int\) cos (ax + b) = \(1\over a\) sin(ax + b) + C.

Solution : We have, 

I = \(\int\) cos (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

\(\int\) cos (ax + b) dx = \(1\over a\) \(\int\) cos t dt

= \(1\over a\) sin t + C

= \(1\over a\) sin(ax + b) + C

Hence, \(\int\) cos (ax + b) = \(1\over a\) sin(ax + b) + C


Related Questions

What is the Differentiation of cos x ?

What is the Integration of Cos Inverse x ?

What is the Differentiation of cos inverse x ?

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