Find \(\int \frac{\sin ^{3} x}{\cos x}+\sin x \cos x \mathrm{~d} x\).

Short Answer

Expert verified
Question: Integrate the following function: \(\int \frac{\sin^3 x}{\cos x}+\sin x \cos x \mathrm{~d} x\) Answer: The integral of the given function is: \(-\cos x + \frac{1}{2}\sin^2 x + C\)

Step by step solution

01

Divide the integrand into two parts

The integrand can be divided into two parts as follows: \[ \frac{\sin^3x}{\cos x} + \sin x \cos x \]
02

Integrate \(\frac{\sin^3x}{\cos x}\) with substitution

Let's use the substitution method to integrate \(\frac{\sin^3x}{\cos x}\). Let \(u = \sin x\), so \(du = \cos x dx\). Then, the first integral can be written as: \[ \int \frac{u^3}{\sqrt{1-u^2}} du \] Next, use a second substitution to solve the integral. Let \(v = 1 - u^2\), so \(dv = -2u du\). We have: \[ -\frac{1}{2} \int \frac{dv}{\sqrt{v}} \] Now, integrate with respect to \(v\): \[ -\frac{1}{2} \cdot 2\sqrt{v} + C_1 = -\sqrt{1 - u^2} + C_1 \] Replace \(u = \sin x\) to find the first part of our solution: \[ -\sqrt{1 - \sin^2 x} + C_1 = -\sqrt{\cos^2 x} + C_1 = -\cos x + C_1 \]
03

Integrate \(\sin x \cos x\)

Now, we need to find the integral of the second part, which is given as \(\sin x \cos x\). Using the substitution method again, let \(w = \sin x\), so \(dw = \cos x dx\). Rewrite the integral as: \[ \int w \, dw \] Then, integrate with respect to \(w\): \[ \frac{1}{2}w^2 + C_2 = \frac{1}{2}\sin^2 x + C_2 \]
04

Add both integrals

Finally, we add the two integrals we found in steps 2 and 3: \[ -\cos x + C_1 + \frac{1}{2}\sin^2 x + C_2 = -\cos x + \frac{1}{2}\sin^2 x + C \] Therefore, the requested integral is: \[ \int \frac{\sin^3 x}{\cos x}+\sin x \cos x \mathrm{~d} x = -\cos x + \frac{1}{2}\sin^2 x + C \]

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