Evaluate the integral. (Use C for the constant of integration.) 8 tan³(x) sec(x) dx 8 sec³(x) 3-sec(x) +C
The Correct Answer and Explanation is :
To evaluate the integral ∫8tan3(x)sec(x) dx\int 8 \tan^3(x) \sec(x) \, dx, let’s break it down step by step.
Step 1: Simplify the expression
We know the relationships between tan(x)\tan(x) and sec(x)\sec(x): sec2(x)=1+tan2(x)\sec^2(x) = 1 + \tan^2(x) ddx[tan(x)]=sec2(x)\frac{d}{dx}[\tan(x)] = \sec^2(x)
This suggests a substitution involving tan(x)\tan(x).
Step 2: Substitution
Let: u=tan(x)u = \tan(x)
Then: du=sec2(x) dxdu = \sec^2(x) \, dx
Also, since sec(x)=1+u2\sec(x) = \sqrt{1 + u^2}, we can rewrite the given integral in terms of uu.
Step 3: Rewrite the integral
The given integral becomes: 8∫tan3(x)sec(x) dx=8∫u31+u2 du8 \int \tan^3(x) \sec(x) \, dx = 8 \int u^3 \sqrt{1 + u^2} \, du
Step 4: Simplify further
Expand the integral: ∫u31+u2 du\int u^3 \sqrt{1 + u^2} \, du
We use substitution to handle the square root. Let: v=1+u2so thatdv=2u duv = 1 + u^2 \quad \text{so that} \quad dv = 2u \, du
Substitute u2=v−1u^2 = v – 1 and u du=12dvu \, du = \frac{1}{2} dv, and the integral becomes: 8∫u31+u2 du=8∫(v−1)v⋅12dv8 \int u^3 \sqrt{1 + u^2} \, du = 8 \int (v – 1) \sqrt{v} \cdot \frac{1}{2} dv
Simplify: 4∫(v3/2−v1/2) dv4 \int (v^{3/2} – v^{1/2}) \, dv
Step 5: Integrate
Use the power rule for each term: ∫v3/2 dv=25v5/2,∫v1/2 dv=23v3/2\int v^{3/2} \, dv = \frac{2}{5} v^{5/2}, \quad \int v^{1/2} \, dv = \frac{2}{3} v^{3/2}
So: 4∫(v3/2−v1/2) dv=4(25v5/2−23v3/2)4 \int (v^{3/2} – v^{1/2}) \, dv = 4 \left( \frac{2}{5} v^{5/2} – \frac{2}{3} v^{3/2} \right)
Step 6: Substitute back
Replace v=1+u2v = 1 + u^2 and u=tan(x)u = \tan(x): =85(1+tan2(x))5/2−83(1+tan2(x))3/2+C= \frac{8}{5} (1 + \tan^2(x))^{5/2} – \frac{8}{3} (1 + \tan^2(x))^{3/2} + C
Final Answer:
The evaluated integral is: 85sec5(x)−83sec3(x)+C\frac{8}{5} \sec^5(x) – \frac{8}{3} \sec^3(x) + C
This matches the expression in terms of sec(x)\sec(x).