MATH 136, section 1 – Calculus 2 Integration By Substitution Practice – Solutions February 7, 2014

Each of the following integrals can be done by the method of u-substitution:

1. Determine the appropriate function u,

2. Compute du,

3. Change the integral to an equivalent form in the new variable u. If it is a definite integral, you can convert the limits of integration as well.

4. Integrate, then

5. Resubstitute u to express the answer in terms of the original variable (indefinite integral cases), or evaluate (definite integral cases).

• ∫ x\sqrt{x^{2} + 16} dx. Let u = x^{2} + 16. Then du = 2x dx and x dx = \frac{1}{2}du.

So the integral is

\frac{1}{2} ∫ \sqrt{u} dx = \frac{1}{2} · \frac{2}{3}u^{3/2} + C = \frac{1}{3}(x^{2} + 16)^{3/2} + C.

• ∫ cos θe^{1+sin θ} dθ. Let u = 1 + sin θ. Then du = cos θ and the integral

becomes:

∫ e^{u} du = e^{u} + C = e^{1+sin θ} + C.

• ∫_{π/4}^{π/2} cos^{3}(4θ) sin(4θ) dθ. Let u = cos(4θ). Then du = −4 sin(4θ) dθ

(chain rule) and sin(4θ) dθ = \frac{−1}{4}du. Also, the limits of integration become x = π/4 ⇒ u = cos(π) = −1 and x = π/2 ⇒ u = cos(2π) = 1. So the integral becomes

\frac{−1}{4} ∫_{−1}^{1} u^{3} du = \frac{−u^{4}}{16} |_{−1}^{1} = 0.

• ∫ \frac{1}{\sqrt{1 − 4x^{2}}} dx. For this one, note the similarity with the formula for

the derivative of the inverse sine function if we think of \sqrt{1 − 4x^{2}} as \sqrt{1 − (2x)^{2}}. If u = 2x, then du = 2 dx so the integral becomes:

\frac{1}{2} ∫ \frac{du}{\sqrt{1 − u^{2}}} = \frac{1}{2} sin^{−1}(u) + C = \frac{1}{2} sin^{−1}(2x) + C.

• ∫_{0}^{1/4} \frac{x}{\sqrt{1 − 4x^{2}}} dx. This one is different from the previous one because

of the extra x in the numerator. That x is part of du for u = 1 − 4x^{2}. With that substitution, the integral becomes

\frac{−1}{8} ∫_{u=1}^{u=3/4} u^{−1/2} du = \frac{−1}{4}u^{1/2}|_{1}^{3/4} = \frac{1}{4} (1 − \sqrt{\frac{3}{4}}) .

• ∫ \frac{1}{x ln(x)} dx. Let u = ln(x), then du = \frac{1}{x} dx and the integral becomes

∫ \frac{1}{u} du = ln |u| + C = ln | ln(x)| + C.

• ∫ \frac{cos(\sqrt{y})}{\sqrt{y}} dy. Let u = \sqrt{y}. Then du = \frac{1}{2\sqrt{y}} dy, so the integral is

2 ∫ cos(u) du = 2 sin(u) + C = 2 sin(\sqrt{y}) + C.