Question

$\int \frac{\sin x + \sin^3 x}{\cos 2x} \, dx = A \cos x + B \log f(x) + c$ (where $c$ is a constant of integration). Then values of $A$, $B$ and $f(x)$ are

• A. $A=\frac{1}{2}$, $B=\frac{-3}{4\sqrt{2}}$, $f(x)=\frac{\sqrt{2}\cos~x-1}{\sqrt{2}\cos~x+1}$
• B. $A=-\frac{1}{2}$, $B=\frac{-3}{4\sqrt{2}}$, $f(x)=\frac{\sqrt{2}\cos~x+1}{\sqrt{2}\cos~x-1}$
• C. $A=\frac{1}{2}$, $B=\frac{-3}{4\sqrt{2}}$, $f(x)=\frac{\sqrt{2}\cos~x+1}{\sqrt{2}\cos~x-1}$
• D. $A=\frac{3}{2}$, $B=\frac{1}{2}$, $f(x)=\frac{\sqrt{2}\cos~x-1}{\sqrt{2}\cos~x+1}$

Hint:

Calculus Tip: When degree of numerator equals the degree of denominator in an integrand fraction, always use algebraic manipulation (long division or adding/subtracting terms) to split it into a polynomial plus a proper fraction before integrating.

Answer 1

The Correct Option is A

Concept: Calculus - Indefinite Integration by Substitution.

Step 1: Express the integrand in terms of $\cos x$. Let $I = \int \frac{\sin x + \sin^3 x}{\cos 2x} dx$. Factor out $\sin x$ in the numerator: $\int \frac{\sin x (1 + \sin^2 x)}{\cos 2x} dx$. Use the identity $\sin^2 x = 1 - \cos^2 x$ for the numerator and $\cos 2x = 2\cos^2 x - 1$ for the denominator. The integral becomes $\int \frac{\sin x (1 + 1 - \cos^2 x)}{2\cos^2 x - 1} dx = \int \frac{\sin x (2 - \cos^2 x)}{2\cos^2 x - 1} dx$.

Step 2: Apply integration by substitution. Let $t = \cos x$. Differentiating both sides gives $dt = -\sin x~dx$, which means $\sin x~dx = -dt$. Substitute this into the integral: $I = \int \frac{-(2 - t^2)}{2t^2 - 1} dt = \int \frac{t^2 - 2}{2t^2 - 1} dt$.

Step 3: Perform algebraic manipulation to simplify the fraction. Multiply and divide the integral by 2 to match the denominator's leading term: $I = \frac{1}{2} \int \frac{2t^2 - 4}{2t^2 - 1} dt$. Now, split the numerator to create a term identical to the denominator: $I = \frac{1}{2} \int \frac{(2t^2 - 1) - 3}{2t^2 - 1} dt = \frac{1}{2} \int \left( 1 - \frac{3}{2t^2 - 1} \right) dt$.

Step 4: Integrate the simplified terms. Separate into two integrals: $\frac{1}{2} \int 1~dt - \frac{3}{2} \int \frac{1}{(\sqrt{2}t)^2 - 1^2} dt$. The first integral is simply $t$. For the second, use the standard formula $\int \frac{1}{x^2 - a^2} dx = \frac{1}{2a} \ln \left| \frac{x-a}{x+a} \right|$. $I = \frac{1}{2}t - \frac{3}{2} \left[ \frac{1}{2(1)\sqrt{2}} \ln \left| \frac{\sqrt{2}t - 1}{\sqrt{2}t + 1} \right| \right] + c$. $I = \frac{1}{2}t - \frac{3}{4\sqrt{2}} \ln \left| \frac{\sqrt{2}t - 1}{\sqrt{2}t + 1} \right| + c$.

Step 5: Substitute back and compare to find the constants. Substitute $t = \cos x$ back into the expression: $I = \frac{1}{2}\cos x - \frac{3}{4\sqrt{2}} \log \left| \frac{\sqrt{2}\cos x - 1}{\sqrt{2}\cos x + 1} \right| + c$. Comparing this with the given format $A\cos x + B\log f(x) + c$, we extract: $A = \frac{1}{2}$, $B = \frac{-3}{4\sqrt{2}}$, and $f(x) = \frac{\sqrt{2}\cos x - 1}{\sqrt{2}\cos x + 1}$. $$ \therefore \text{The correct match is Option A.} $$