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

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

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