Question

Integration of 1 -cos square x .sinx

Answer 1

(cos^3 x)/3 - cos x + C

Answer 2

The given expression to integrate is $(1 - \cos^2 x) \cdot \sin x$.

First, simplify the expression using the trigonometric identity $\sin^2 x + \cos^2 x = 1$.
From this, we can write $1 - \cos^2 x = \sin^2 x$.

Substitute this into the expression:
$(1 - \cos^2 x) \cdot \sin x = \sin^2 x \cdot \sin x = \sin^3 x$.

Now, we need to find the integral of $\sin^3 x$.
To integrate $\sin^3 x$, we can rewrite it as $\sin^2 x \cdot \sin x$.
Again, use the identity $\sin^2 x = 1 - \cos^2 x$.

So the integral becomes:

$$\int \sin^3 x \, dx = \int (1 - \cos^2 x) \sin x \, dx$$

Now, use the method of substitution.
Let $u = \cos x$.
Then, differentiate both sides with respect to $x$:

$$\frac{du}{dx} = -\sin x$$

This implies $du = -\sin x \, dx$, or $\sin x \, dx = -du$.

Substitute $u$ and $du$ into the integral:

$$\int (1 - u^2) (-du)$$ $$= - \int (1 - u^2) \, du$$

Now, integrate term by term:

$$= - \left( \int 1 \, du - \int u^2 \, du \right)$$ $$= - \left( u - \frac{u^3}{3} \right) + C$$ $$= -u + \frac{u^3}{3} + C$$

Finally, substitute back $u = \cos x$:

$$= -\cos x + \frac{\cos^3 x}{3} + C$$

The constant of integration $C$ is added because it is an indefinite integral.