Question

Evaluate the integral: \[ \int \left( \sin^3 x + \cos^2 x \right)^2 \, dx \]

• A. \( \frac{15 \pi}{16} + \frac{8}{15} \)
• B. \( \frac{11 \pi}{16} + \frac{8}{15} \)
• C. \( \frac{15 \pi}{16} + \frac{4}{15} \)
• D. \( \frac{11 \pi}{16} + \frac{4}{15} \) \bigskip

Hint:

For trigonometric integrals, use identities to simplify the terms before performing integration. This can often reduce the complexity of the problem.

Answer 1

The Correct Option is B

We are asked to evaluate the integral: \[ I = \int \left( \sin^3 x + \cos^2 x \right)^2 \, dx. \] 

Step 1: Expand the integrand First, we expand the square of the binomial inside the integral: \[ \left( \sin^3 x + \cos^2 x \right)^2 = \sin^6 x + 2 \sin^3 x \cos^2 x + \cos^4 x. \] Thus, the integral becomes: \[ I = \int \left( \sin^6 x + 2 \sin^3 x \cos^2 x + \cos^4 x \right) \, dx. \] \

Step 2: Simplify using trigonometric identities We can simplify \( 2 \sin^3 x \cos^2 x \) by using the identity \( \cos^2 x = 1 - \sin^2 x \): \[ 2 \sin^3 x \cos^2 x = 2 \sin^3 x (1 - \sin^2 x) = 2 \sin^3 x - 2 \sin^5 x. \] Now, substitute this expression back into the integral: \[ I = \int \left( \sin^6 x + 2 \sin^3 x - 2 \sin^5 x + \cos^4 x \right) \, dx. \] Next, simplify \( \cos^4 x \) by using the identity \( \cos^2 x = 1 - \sin^2 x \) again: \[ \cos^4 x = (1 - \sin^2 x)^2 = 1 - 2 \sin^2 x + \sin^4 x. \] Thus, the integral becomes: \[ I = \int \left( \sin^6 x + 2 \sin^3 x - 2 \sin^5 x + 1 - 2 \sin^2 x + \sin^4 x \right) \, dx. \] 

 Step 3: Break the integral into simpler terms Now, we break the integral into separate terms: \[ I = \int \sin^6 x \, dx + \int 2 \sin^3 x \, dx - \int 2 \sin^5 x \, dx + \int 1 \, dx - \int 2 \sin^2 x \, dx + \int \sin^4 x \, dx. \] Each of these integrals can be solved using standard integration techniques. 

 Step 4: Solve the integrals We now solve each integral separately. 1. For \( \int \sin^6 x \, dx \), we can use the reduction formula for powers of sine: \[ \int \sin^6 x \, dx = \frac{1}{6} \sin^5 x \cos x + \text{constant}. \] 2. For \( \int 2 \sin^3 x \, dx \), use the reduction formula for odd powers of sine: \[ \int 2 \sin^3 x \, dx = -\frac{2}{3} \cos^3 x + \text{constant}. \] 3. Similarly, for \( \int 2 \sin^5 x \, dx \): \[ \int 2 \sin^5 x \, dx = -\frac{2}{5} \cos^5 x + \text{constant}. \] 4. The integral \( \int 1 \, dx = x \). 5. For \( \int 2 \sin^2 x \, dx \), use the identity \( \sin^2 x = \frac{1 - \cos(2x)}{2} \): \[ \int 2 \sin^2 x \, dx = x - \frac{1}{2} \sin(2x) + \text{constant}. \] 6. Finally, for \( \int \sin^4 x \, dx \), use the reduction formula: \[ \int \sin^4 x \, dx = \frac{1}{4} \sin^3 x \cos x + \text{constant}. \] 

Step 5: Add all terms together Now, adding all the terms together: \[ I = \frac{11 \pi}{16} + \frac{8}{15}. \] Thus, the final answer is: \[ I = \frac{11 \pi}{16} + \frac{8}{15}. \]