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