Question

Evaluate the following integrals: (i) \(\int \frac{1+\cos ^{2} x}{1+\cos 2 x} d x\) (ii) \(\int \frac{1-\tan ^{2} x}{1+\tan ^{2} x} d x\) (iii) \(\int \frac{1+\tan ^{2} x}{1+\cot ^{2} x} d x\) (iv) \(\int \frac{\cos 2 x}{\cos ^{2} x \sin ^{2} x} d x\)

Short answer

In this exercise, we evaluated four different integrals using various integration techniques like substitution, trigonometric identities, and integrating by parts. The results are as follows: (i) \(\int \frac{1+\cos ^{2} x}{1+\cos 2 x} dx = \frac{1}{2}[ \tan x+x ]+C\) (ii) \(\int \frac{1-\tan ^{2} x}{1+\tan ^{2} x} dx = \frac{1}{2}\sin 2x + C\) (iii) \(\int \frac{1+\tan ^{2} x}{1+\cot ^{2} x} dx = x - \ln|1-\sin^2x| + C\) (iv) \(-\int \frac{\cos 2x}{\cos^2x\sin^2x} dx = -(\cos x + \ln|1-\cos x|-\ln|1+\cos x|+C)\)

Step-by-step solution

Integral (i) \(\int \frac{1+\cos ^{2} x}{1+\cos 2 x} d x\)

To solve this integral, we'll use the trigonometric identity \(\cos 2x = 2\cos^2x - 1\). We can rewrite the integral as: \(\int \frac{1 + \cos^2x}{1 + 2\cos^2x - 1} dx\) Now the integral becomes: \(\int \frac{1+\cos^2x}{2\cos^2x} dx = \frac{1}{2}\int\frac{1+\cos^2x}{\cos^2x} dx\) Now split the fraction: \(\frac{1}{2}\int\left(\frac{1}{\cos^2x}+\frac{\cos^2x}{\cos^2x}\right) dx = \frac{1}{2}\int\left(\frac{1}{\cos^2x}+1\right) dx\) Use the identity \(\sec^2x = \frac{1}{\cos^2x}\): \(\frac{1}{2}\int\left(\sec^2x+1\right) dx\) Now integrate: \(\frac{1}{2} \left[\int \sec^2x dx + \int 1 dx\right] = \frac{1}{2}[ \tan x+x ]+C\) So, the final result is: \(\int \frac{1+\cos ^{2} x}{1+\cos 2 x} d x = \frac{1}{2}[ \tan x+x ]+C\)

Integral (ii) \(\int \frac{1-\tan ^{2} x}{1+\tan ^{2} x} d x\)

To solve this integral, we can first notice that the integrand is similar to the trigonometric identity \(\cos 2x = \frac{1-\tan^2x}{1+\tan^2x}\). Rewrite the integral as: \(\int \cos 2x dx\) Now integrate: \(\int \cos 2x dx = \frac{1}{2}\sin 2x + C\) The result is: \(\int \frac{1-\tan ^{2} x}{1+\tan ^{2} x} d x = \frac{1}{2}\sin 2x + C\)

Integral (iii) \(\int \frac{1+\tan ^{2} x}{1+\cot ^{2} x} d x\)

For this integral, we can rewrite it in terms of sines and cosines: \(\int \frac{1+\frac{\sin^2x}{\cos^2x}}{1+\frac{\cos^2x}{\sin^2x}} dx\) Now, find a common denominator for both fractions: \(\int \frac{\frac{\sin^2x+\cos^2x}{\cos^2x}}{\frac{\sin^2x+\cos^2x}{\sin^2x}} dx\) The integrand simplifies to: \(\int \frac{\sin^2x}{\cos^2x} dx\) Now use the substitution \(u = \sin x\) and \(du = \cos x dx\): \(\int \frac{u^2}{1-u^2} du\) Now perform long division, which results in: \(\int\left(1+\frac{u^2}{1-u^2}\right) du\) Next, integrate: \(\int 1 du + \int\frac{u^2}{1-u^2} du = u -\int \frac{1-u^2+u^2}{1-u^2} du\) Now integrate the remaining term: \(u - \int \frac{1}{1-u^2} du = u - \ln|1-u^2| + C = x - \ln|1-\sin^2x| + C\) Thus, the final result is: \(\int \frac{1+\tan ^{2} x}{1+\cot ^{2} x} d x = x - \ln|1-\sin^2x| + C\)

Integral (iv) \(\int \frac{\cos 2x}{\cos^2x\sin^2x} dx\)

First, use the identity \(\cos 2x = 2\cos^2x - 1\) and then \(\sin^2x = 1-\cos^2x\): \(\int \frac{2\cos^2x-1}{\cos^2x (1-\cos^2x)} dx\) Now, integrate: \(\int \frac{2\cos^2x-1}{(1-\cos^2x)} dx\) Use the substitution \(u = \cos x\) and \(du = -\sin x dx\): \(-\int\frac{2u^2-1}{1-u^2} du\) Now, perform partial fractions: \(-\int\left(1-\frac{1}{1-u}+\frac{1}{1+u}\right) du = -(u+\ln|1-u|-\ln|1+u|+C)\) Now, resubstitute back: \(-\int \frac{\cos 2x}{\cos^2x\sin^2x} dx = -(\cos x + \ln|1-\cos x|-\ln|1+\cos x|+C)\).