Q13. Compute the integral: integral(sin2 x / (1 + cos x)) dx

Q10. Evaluate the integral: $\int \left(\frac{x}{x+1}\right) dx$

Step 1: Rewrite the integrand using algebraic manipulation. $\frac{x}{x+1} = \frac{x+1-1}{x+1} = 1 - \frac{1}{x+1}$ Step 2: Integrate each term. $\int \left(1 - \frac{1}{x+1}\right) dx = \int 1 \, dx - \int \frac{1}{x+1} \, dx$ Step 3: Perform the integration. $\int 1 \, dx = x$ $\int \frac{1}{x+1} \, dx = \ln|x+1|$ Step 4: Combine the results and add the constant of integration. $x - \ln|x+1| + C$ The final answer is $\boxed{x - \ln|x+1| + C}$.

Q11. Evaluate the integral: $\int \tan^2 x \, dx$

Step 1: Use the trigonometric identity $\tan^2 x = \sec^2 x - 1$. $\int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx$ Step 2: Integrate each term. $\int \sec^2 x \, dx - \int 1 \, dx$ Step 3: Perform the integration. $\int \sec^2 x \, dx = \tan x$ $\int 1 \, dx = x$ Step 4: Combine the results and add the constant of integration. $\tan x - x + C$ The final answer is $\boxed{\tan x - x + C}$.

Q12. Compute the indefinite integral: $\int \sqrt{1-\sin 2x} \, dx$

Step 1: Use the identities $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$ to rewrite the expression under the square root. $1 - \sin 2x = \sin^2 x + \cos^2 x - 2 \sin x \cos x = (\cos x - \sin x)^2$ Step 2: Substitute this back into the integral. $\int \sqrt{(\cos x - \sin x)^2} \, dx = \int |\cos x - \sin x| \, dx$ Step 3: For simplicity, assume $\cos x - \sin x \ge 0$ over the interval of integration, allowing us to remove the absolute value. $\int (\cos x - \sin x) \, dx$ Step 4: Integrate each term. $\int \cos x \, dx - \int \sin x \, dx$ Step 5: Perform the integration. $\int \cos x \, dx = \sin x$ $\int \sin x \, dx = -\cos x$ Step 6: Combine the results and add the constant of integration. $\sin x - (-\cos x) + C = \sin x + \cos x + C$ The final answer is $\boxed{\sin x + \cos x + C}$.

Q13. Compute the integral: $\int \frac{\sin^2 x}{1+\cos x} \, dx$

Step 1: Use the trigonometric identity $\sin^2 x = 1 - \cos^2 x$. $\int \frac{1 - \cos^2 x}{1+\cos x} \, dx$ Step 2: Factor the numerator using the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$. $1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$ Step 3: Substitute the factored numerator back into the integral and simplify. $\int \frac{(1 - \cos x)(1 + \cos x)}{1+\cos x} \, dx = \int (1 - \cos x) \, dx$ Step 4: Integrate each term. $\int 1 \, dx - \int \cos x \, dx$ Step 5: Perform the integration. $\int 1 \, dx = x$ $\int \cos x \, dx = \sin x$ Step 6: Combine the results and add the constant of integration. $x - \sin x + C$ The final answer is $\boxed{x - \sin x + C}$.

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