# Find the integral of cos4 (x) dx

We have to find the integral of cos4x dx

### Solution

We know that cos4x can be written as cos3x .dx

Now will integrate cos3x .dx by parts

$\int \cos ^{4}x.dx=\int \cos ^{3}x \cos x.dx$ $\int \cos ^{4}x.dx=\int \cos ^{3}x d(\sin x)$ $\int \cos ^{4}x.dx=\sin x\cos ^{3}x – \int \sin x d(\cos ^{3}x)dx$ $\int \cos ^{4}x.dx=\sin x\cos ^{3}x + 3 \int \sin^{2} \cos ^{2}xdx$

Now we will use the identity

sin2x = 1-cos2x

$\int \cos ^{4}x.dx=\sin x\cos ^{3}x + 3 \int (1-sin^{2}x) \cos ^{2}xdx$ $\int \cos ^{4}x.dx=\sin x\cos ^{3}x + 3 \int (cos^{2}x) -3\int \cos ^{2}xdx$

We have now the same integral on both sides and we can solve for it:

$4\int \cos ^{4}x.dx=\sin x\cos ^{3}x + 3 \int (cos^{2}x)dx$ $\int \cos ^{4}x.dx=\frac{\sin x\cos ^{3}x}{4} + \frac{3}{4} \int (cos^{2}x)dx$

Using the same process we get

$\int \cos ^{2}x.dx=\int \cos x d(sin x)= cos x sin x +\int sin^{2}x dx$ $\int \cos ^{2}x.dx=\int \cos x d(sin x)= cos x sin x +\int 1 – cos^{2}x dx$ $\int \cos ^{2}x.dx=\int \cos x d(sin x)= cos x sin x + x -\int cos^{2}x dx$ $\int \cos ^{2}x.dx=\frac{ cos x sin x}{2} + \frac{x}{2}$

Substituting in the expression above we get

$\int \cos ^{4}x.dx=\frac{sinx cos^{3}x}{4} +\frac{3}{8}(cosx sin x) +\frac{3}{8}x$

$\int \cos ^{4}x.dx=\frac{sinx cos^{3}x}{4} +\frac{3}{8}(cosx sin x) +\frac{3}{8}x$