Evaluate the following integrals: a) ( int cos ^{4}(3 x) sin (3 x) d x ) b) ( int_{0}^{2} x e^{2 x} d x ) c) ( int_{1}^{4} frac{2 x}{3+x^{2}} d x )
Added by Ella P.
Close
Step 1
Let \( u = \cos(3x) \), then \( du = -3\sin(3x)dx \). The integral becomes \( -\frac{1}{3} \int u^4 du \), which can be solved as \( -\frac{1}{3} \cdot \frac{u^5}{5} + C = -\frac{1}{15} \cos^5(3x) + C \). Show more…
Show all steps
Your feedback will help us improve your experience
Israel Hernandez and 91 other Calculus 2 / BC educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
evaluate
Ekaveera K.
William S.
Evaluate the integrals by making the indicated substitutions. (a) $\int 2 x\left(x^{2}+1\right)^{23} d x ; u=x^{2}+1$ (b) $\int \cos ^{3} x \sin x d x ; u=\cos x$ (c) $\int \frac{1}{\sqrt{x}} \sin \sqrt{x} d x ; u=\sqrt{x}$ (d) $\int \frac{3 x d x}{\sqrt{4 x^{2}+5}} ; u=4 x^{2}+5$ (e) $\int \frac{x^{2}}{x^{3}-4} d x ; u=x^{3}-4$
Integration
Integration by Substitution
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD