# Thread: Integration of Exponential and Trignometric functions

1. ## Integration of Exponential and Trignometric functions

Let S be the integration symbol

1. S (dx)/ (x. sec ^3 (ln (x)) dx
2. S (x^3)/(sqrt (x^2 -4 )) dx
3. S (2sec x)/(sin x + cos x) dx
4. S sqrt (sqrt (x) -1) dx
5. S sec^3 (x). tan^3 (x) dx
6. S sin (5x)*cos (7x) dx
7. S (e^(x) -3)/(e^(x) +2) dx
8. S (sec^4 (x))/(tan^2 (x)) dx
9. S (sin 2x)/(1 + 3sin^2 (x))^3 dx

Sorry for the many questions, but I wanna make sure of everything related to Integration before my Exam the next Thursday.

2. 6. We have for $a\in \mathbb R$ and $b\in \mathbb R$ $\sin\left(a+b\right) = \sin a \cos b+\sin b\cos a$ and $\sin\left(a-b\right) = \sin a \cos b-\sin b\cos a$ so
$\sin \left(a+b\right)-\sin \left(a-b\right) = 2\sin b\cos a$.
We apply this result for $a=5x$ and $b=7x$:
$I:=\int \cos\left(5x\right)\sin\left(7x\right)\, dx =\int \frac 12\left(\sin \left(12 x\right)-\sin\left(-2x\right)\right) \, dx$
$I= \frac 12 \int \sin \left(12x\right)\, dx +\frac 12 \int \sin \left(2x\right)\,dx= -\frac 1{24} \cos\left(12x\right)-\frac 14\cos\left(2x\right)$

7. $I:=\int \frac{e^x-3}{e^x+2}\,dx= \int\left(1-\frac 5{e^x+2}\right)\, dx
=x-5\int \frac{e^{-x}}{1+2e^{-x}}\,dx
$
so
$I= x-\frac 52 \int\frac{2e^{-x}}{1+2e^{-x}}\, dx$ and finally $I = x+\frac 52 \ln\left(1+2e^{-x}\right)$

3. Originally Posted by egt2010
Let S be the integration symbol

1. S (dx)/ (x. sec ^3 (ln (x)) dx
2. S (x^3)/(sqrt (x^2 -4 )) dx
3. S (2sec x)/(sin x + cos x) dx
4. S sqrt (sqrt (x) -1) dx
5. S sec^3 (x). tan^3 (x) dx
6. S sin (5x)*cos (7x) dx
7. S (e^(x) -3)/(e^(x) +2) dx
8. S (sec^4 (x))/(tan^2 (x)) dx
9. S (sin 2x)/(1 + 3sin^2 (x))^3 dx

Sorry for the many questions, but I wanna make sure of everything related to Integration before my Exam the next Thursday.

It is completely unreasonable that you would expect this list of questions to get answered. Especially since there is no sign of any attempt or effort having been made by you.

Use this website (be sure to click on Show steps after submitting the question): Wolfram|Alpha

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