# 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 ?