Question

\(\int\sin^3xdx+\int\cos^2x\sin xdx=\)

• A. -cosx+C
• B. -sinx+C
• C. x-cosx+C
• D. x-sinx+C
• E. cosx-sinx+C

Answer 1

The Correct Option is A

We are tasked with evaluating the integral: \[ \int \sin^3 x \, dx + \int \cos^2 x \sin x \, dx. \]
Step 1: Break the problem into two integrals We can split the given problem into two separate integrals: \[ \int \sin^3 x \, dx + \int \cos^2 x \sin x \, dx. \]
Step 2: Solve the first integral \( \int \sin^3 x \, dx \) To solve \( \int \sin^3 x \, dx \), use the identity \( \sin^3 x = \sin x (1 - \cos^2 x) \), and break it into simpler terms: \[ \int \sin^3 x \, dx = \int \sin x (1 - \cos^2 x) \, dx. \] Now, let \( u = \cos x \), so that \( du = -\sin x \, dx \). The integral becomes: \[ \int \sin x (1 - \cos^2 x) \, dx = - \int (1 - u^2) \, du = - \left( u - \frac{u^3}{3} \right) + C = - \cos x + \frac{\cos^3 x}{3} + C_1. \]
Step 3: Solve the second integral \( \int \cos^2 x \sin x \, dx \) Next, solve \( \int \cos^2 x \sin x \, dx \). Use substitution: Let \( u = \cos x \), so that \( du = -\sin x \, dx \). The integral becomes: \[ \int \cos^2 x \sin x \, dx = - \int u^2 \, du = - \frac{u^3}{3} + C_2 = - \frac{\cos^3 x}{3} + C_2. \]
Step 4: Combine the results Now, adding the results of both integrals, we get: \[ - \cos x + \frac{\cos^3 x}{3} - \frac{\cos^3 x}{3} + C = - \cos x + C. \]

The correct option is (A) : \(-cosx+C\)

Answer 2

We want to evaluate the integral: \[\int \sin^3 x \, dx + \int \cos^2 x \sin x \, dx\]

First, let's evaluate \(\int \sin^3 x \, dx\). We can write \(\sin^3 x = \sin^2 x \cdot \sin x = (1 - \cos^2 x) \sin x\). So, \[\int \sin^3 x \, dx = \int (1 - \cos^2 x) \sin x \, dx\]

Let \(u = \cos x\), then \(du = -\sin x \, dx\), so \(\sin x \, dx = -du\). Substituting these into the integral gives: \[\int (1 - u^2) (-du) = -\int (1 - u^2) \, du = \int (u^2 - 1) \, du = \frac{u^3}{3} - u + C_1\] \[\int \sin^3 x \, dx = \frac{\cos^3 x}{3} - \cos x + C_1\]

Now, let's evaluate \(\int \cos^2 x \sin x \, dx\). Using the same substitution \(u = \cos x\) and \(du = -\sin x \, dx\), we get: \[\int \cos^2 x \sin x \, dx = \int u^2 (-du) = -\int u^2 \, du = -\frac{u^3}{3} + C_2\] \[\int \cos^2 x \sin x \, dx = -\frac{\cos^3 x}{3} + C_2\]

Adding the two integrals: \[\int \sin^3 x \, dx + \int \cos^2 x \sin x \, dx = \left(\frac{\cos^3 x}{3} - \cos x + C_1\right) + \left(-\frac{\cos^3 x}{3} + C_2\right) = -\cos x + C\] where \(C = C_1 + C_2\).

Therefore, the integral is: \[\int \sin^3 x \, dx + \int \cos^2 x \sin x \, dx = -\cos x + C\]