Question

$\int \frac{1+cos^2x}{1+cos^2 2x} dx$

Answer 1

$\frac{3\sqrt{2}}{8} \tan^{-1}\left(\frac{\tan 2x}{\sqrt{2}}\right) + \frac{\sqrt{2}}{16} \ln\left|\frac{\sqrt{2}+\sin 2x}{\sqrt{2}-\sin 2x}\right| + C$

Answer 2

The integral to solve is $\int \frac{1+\cos^2x}{1+\cos^2 2x} dx$.

Step 1: Simplify the numerator using trigonometric identities. We know that $\cos^2 x = \frac{1+\cos 2x}{2}$. Substitute this into the numerator: $1+\cos^2 x = 1 + \frac{1+\cos 2x}{2} = \frac{2+1+\cos 2x}{2} = \frac{3+\cos 2x}{2}$.

Step 2: Rewrite the integral. The integral becomes: $\int \frac{\frac{3+\cos 2x}{2}}{1+\cos^2 2x} dx = \frac{1}{2} \int \frac{3+\cos 2x}{1+\cos^2 2x} dx$.

Step 3: Use substitution to simplify the argument of trigonometric functions. Let $u = 2x$. Then $du = 2dx$, which implies $dx = \frac{1}{2}du$. Substitute $u$ and $dx$ into the integral: $\frac{1}{2} \int \frac{3+\cos u}{1+\cos^2 u} \frac{1}{2} du = \frac{1}{4} \int \frac{3+\cos u}{1+\cos^2 u} du$.

Step 4: Split the integral into two parts. $\frac{1}{4} \int \left( \frac{3}{1+\cos^2 u} + \frac{\cos u}{1+\cos^2 u} \right) du$.

Step 5: Evaluate the first part of the integral. Let $I_1 = \int \frac{3}{1+\cos^2 u} du$. Divide the numerator and denominator by $\cos^2 u$: $I_1 = \int \frac{3/\cos^2 u}{1/\cos^2 u + 1} du = \int \frac{3\sec^2 u}{\sec^2 u + 1} du$. Use the identity $\sec^2 u = 1+\tan^2 u$: $I_1 = \int \frac{3\sec^2 u}{(1+\tan^2 u) + 1} du = \int \frac{3\sec^2 u}{2+\tan^2 u} du$. Now, let $t = \tan u$. Then $dt = \sec^2 u \, du$. $I_1 = \int \frac{3 dt}{2+t^2} = 3 \int \frac{dt}{(\sqrt{2})^2+t^2}$. This is a standard integral of the form $\int \frac{dx}{a^2+x^2} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)$. $I_1 = 3 \cdot \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t}{\sqrt{2}}\right) = \frac{3}{\sqrt{2}} \tan^{-1}\left(\frac{\tan u}{\sqrt{2}}\right)$.

Step 6: Evaluate the second part of the integral. Let $I_2 = \int \frac{\cos u}{1+\cos^2 u} du$. Let $v = \sin u$. Then $dv = \cos u \, du$. Also, $\cos^2 u = 1-\sin^2 u = 1-v^2$. Substitute these into the integral: $I_2 = \int \frac{dv}{1+(1-v^2)} = \int \frac{dv}{2-v^2}$. This is a standard integral of the form $\int \frac{dx}{a^2-x^2} = \frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right|$. Here $a=\sqrt{2}$. $I_2 = \frac{1}{2\sqrt{2}} \ln\left|\frac{\sqrt{2}+v}{\sqrt{2}-v}\right|$. Substitute back $v = \sin u$: $I_2 = \frac{1}{2\sqrt{2}} \ln\left|\frac{\sqrt{2}+\sin u}{\sqrt{2}-\sin u}\right|$.

Step 7: Combine the results and substitute back $u=2x$. The total integral is $\frac{1}{4} (I_1 + I_2)$. $\int \frac{1+\cos^2x}{1+\cos^2 2x} dx = \frac{1}{4} \left[ \frac{3}{\sqrt{2}} \tan^{-1}\left(\frac{\tan u}{\sqrt{2}}\right) + \frac{1}{2\sqrt{2}} \ln\left|\frac{\sqrt{2}+\sin u}{\sqrt{2}-\sin u}\right| \right] + C$. Substitute $u=2x$: $= \frac{1}{4} \left[ \frac{3}{\sqrt{2}} \tan^{-1}\left(\frac{\tan 2x}{\sqrt{2}}\right) + \frac{1}{2\sqrt{2}} \ln\left|\frac{\sqrt{2}+\sin 2x}{\sqrt{2}-\sin 2x}\right| \right] + C$. Simplify the coefficients: $= \frac{3}{4\sqrt{2}} \tan^{-1}\left(\frac{\tan 2x}{\sqrt{2}}\right) + \frac{1}{8\sqrt{2}} \ln\left|\frac{\sqrt{2}+\sin 2x}{\sqrt{2}-\sin 2x}\right| + C$. Rationalize the denominators (optional, but often preferred): $= \frac{3\sqrt{2}}{8} \tan^{-1}\left(\frac{\tan 2x}{\sqrt{2}}\right) + \frac{\sqrt{2}}{16} \ln\left|\frac{\sqrt{2}+\sin 2x}{\sqrt{2}-\sin 2x}\right| + C$.

The final answer is $\frac{3\sqrt{2}}{8} \tan^{-1}\left(\frac{\tan 2x}{\sqrt{2}}\right) + \frac{\sqrt{2}}{16} \ln\left|\frac{\sqrt{2}+\sin 2x}{\sqrt{2}-\sin 2x}\right| + C$.