Question

5.5 EXERCISES 1-6 Evaluate the integral by making the given substitution. 1. ? e^{-x} dx, u = -x 2. ? x^3(2 + x^4)^5 dx, u = 2 + x^4 3. ? x^2?(x^3 + 1) dx, u = x^3 + 1 4. ? dt / (1 - 6t)^4, u = 1 - 6t 5. ? cos^3 ? sin ? d?, u = cos ? 6. ? (sec^2(1/x) / x^2) dx, u = 1/x

          5.5 EXERCISES
1-6 Evaluate the integral by making the given substitution.
1. ? e^{-x} dx, u = -x
2. ? x^3(2 + x^4)^5 dx, u = 2 + x^4
3. ? x^2?(x^3 + 1) dx, u = x^3 + 1
4. ? dt / (1 - 6t)^4, u = 1 - 6t
5. ? cos^3 ? sin ? d?, u = cos ?
6. ? (sec^2(1/x) / x^2) dx, u = 1/x

5.5 EXERCISES
1-6 Evaluate the integral by making the given substitution.
1. ? e^-x dx, u = -x
2. ? x^3(2 + x^4)^5 dx, u = 2 + x^4
3. ? x^2?(x^3 + 1) dx, u = x^3 + 1
4. ? dt / (1 - 6t)^4, u = 1 - 6t
5. ? cos^3 ? sin ? d?, u = cos ?
6. ? (sec^2(1/x) / x^2) dx, u = 1/x

Added by Esther M.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

04/27/2023

Step 1

Now, we need to find $du$ in terms of $d\theta$. Differentiating $u$ with respect to $\theta$, we get: $du = -\sin{\theta} d\theta$ Now, we can rewrite the integral in terms of $u$: $\int \cos^2{\theta} \sin{\theta} d\theta = -\int u^2 du$ Now, we can Show more…

Show all steps

Thanks for your feedback!

5.5 EXERCISES 1-6 Evaluate the integral by making the given substitution: 1. ∫ e^{-x} dx, u = -x 2. ∫ x^3(2 + x^4)^5 dx, u = 2 + x^4 3. ∫ x^2√(x^3 + 1) dx, u = x^3 + 1 4. ∫ dt / (1 - 6t)^4, u = 1 - 6t 5. ∫ cos^3 ̈θ sin θ dθ, u = cos θ 6. ∫ sec^2(1/x) / x^2 dx, u = 1/x