Question
\text { If } I_{n}=\int_{0}^{\pi / 4} \tan ^{n} \theta d \theta, \text { then } n\left(l_{n-1}+i_{m+1}\right)=\ldots
Step 1
Step 1: We are given the integral I_n = ∫(tan^n(θ) dθ) from 0 to π/4. Show more…
Show all steps
Your feedback will help us improve your experience
Aman Gupta and 87 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
If $I_{1, \mathrm{n}}=\int_{0}^{\pi / 2} \frac{\sin (2 n-1) x}{\sin x} d x$ and $I_{2, \mathrm{n}}=\int_{0}^{\pi / 2} \frac{\sin ^{2} n x}{\sin ^{2} x} d x$, $n \in N$, then (A) $I_{2, n+1}-I_{2, n}=I_{1, n}$ (B) $I_{2, n+1}-I_{2, n}=I_{1, n+1}$ (C) $I_{2, n+1}+I_{1, n}=I_{2, n}$ (D) $I_{2, n+1}+I_{1, n+1}=I_{2, n}$
If $I_{n}=\int_{0} e^{-x} \sin ^{n} x d x$, show that $I_{n}=\frac{n(n-1)}{n^{2}+1} I_{m-2}$.
Reduction formulas
Further problems
If Im $n=\int \frac{\sin ^{\mathrm{m}} x}{\cos ^{\prime \prime} x} d x$, then $\operatorname{Im} n=\frac{\sin ^{\prime \prime \prime-1} x}{(n-1) \cos ^{n-1} x}+k \operatorname{lm}_{-2,} n_{-2}$, where $k=$ (A) $\frac{m-1}{n-1}$ (B) $\frac{1-m}{n-1}$ (C) $\frac{m-1}{n}$ (D) $\frac{m}{n-1}$
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD