$$\int_1^2 \operatorname{cosec}^2 4 t\;dt$$
Been trying to evalute this definite integral, although I can't find what $\operatorname{cosec}^2$ is?
Asked
Active
Viewed 105 times
0
-
do you meant $$\int_{1}^{2}\ csc^2(4x)dx$$? – Dr. Sonnhard Graubner Aug 13 '15 at 12:21
-
Yes, I wasn't sure how to make the symbol – joe Aug 13 '15 at 12:22
-
For future reference, @joe, you need to use backslashes not forward slashes when typing these commands. – not all wrong Aug 13 '15 at 12:23
-
$\int_1^2$ is this how? – joe Aug 13 '15 at 12:24
-
or $\operatorname{cosec}{x}=$ – Dr. Sonnhard Graubner Aug 13 '15 at 12:24
-
Also, $\operatorname{cosec} x \equiv \frac 1 {\sin x} $. – not all wrong Aug 13 '15 at 12:26
2 Answers
0
Notice, $$\int_{1}^{2}\csc^24t dt$$ Let, $4t=u\iff 4dt=du\iff dt=\frac{du}{4}$
Now, we have $$\int_{4}^{8}\csc^2(u) \frac{du}{4}$$ $$=\frac{1}{4}\int_{4}^{8}\csc^2(u) du$$
$$=\frac{1}{4}\left[-\cot u\right]_{4}^{8}$$ $$=\frac{1}{4}\left[-\cot 8+\cot 4\right]$$ $$=\frac{1}{4}\left[\cot 4-\cot 8\right]$$
Harish Chandra Rajpoot
- 37,450
-
-
Yes you are right, Since $$\frac{d}{dx}(\cot x)=-\csc^2 x$$ – Harish Chandra Rajpoot Aug 13 '15 at 12:24
-
and the 1/4? probably asking really obvious questions here, i'm a bit rusty on this – joe Aug 13 '15 at 12:29
-
OK, you can let $4t=x\iff 4dt=dx\iff dt=\frac{1}{4}dx$ Substitute this & find the integral you will get the same results. – Harish Chandra Rajpoot Aug 13 '15 at 12:33
0
$$\int\csc^{2n+2}\theta\ d\theta=\int\text{cosec}^{2n+2}\theta\ d\theta=\int(1+\cot^2\theta)^n\ \text{cosec}^2\theta\ d\theta$$
Set $\cot\theta=y$
lab bhattacharjee
- 274,582