Math Help - A few integration problems

1. A few integration problems

$1. \int \sin^3(5x)\cos(5x)dx$

$2. \int \tan^2(x)dx$

$3. \int \frac{\sec^2(x)}{3+\tan(x)}dx$

$4. \int \sin(x)(\cos(x)+\csc(x))dx$

These are what I got:
1. $-\frac{5}{2}\cos^2(5x)+\frac{15}{4}\cos^4(5x)+c$

2. $\tan(x)-x+c$

3. $Ln(3+\tan(x))+c$

Number 4 I can't solve, can someone help? Thanks

2. Originally Posted by nirva

$1. \int \sin^3(5x)\cos(5x)dx$

These are what I got:
1. $-\frac{5}{2}\cos^2(5x)+\frac{15}{4}\cos^4(5x)+c$
Observe that $5 \cos(5x)$ is the derivative of $\sin(5x)$ suggests that the integral is for some $K$:

$
\int \sin^3(5x)\cos(5x)dx = K \sin^4(5x) + c
$

and differentiating the RHS tells us that $K=1/20$ so:

$
\int \sin^3(5x)\cos(5x)dx = 1/20 \sin^4(5x) + c
$

RonL

3. Originally Posted by nirva

...
$4. \int \sin(x)(\cos(x)+\csc(x))dx$
...
Number 4 I can't solve, can someone help? Thanks
Hello, nirva,

$\int \sin(x)(\cos(x)+\csc(x))dx= \int \left( \sin(x)\cdot \cos(x) + \sin(x) \cdot \frac{1}{\sin(x)} \right)dx$

Remember that cos(x) is the derivative of sin(x). Then use substitution:

$\int \left( \sin(x)\cdot \cos(x) + \sin(x) \cdot \frac{1}{\sin(x)} \right)dx= \int \sin(x)\cdot \cos(x) dx+ \int \frac{\sin(x)}{\sin(x)} dx$

(For confirmation only: I've got 1/2*sin^2(x) + x)

EB

4. Hello, nirva!

$1)\;\int \sin^35x\cos5x\,dx$

Let $u = \sin5x\quad\Rightarrow\quad du = 5\cos5x\,dx\quad\Rightarrow\quad dx = \frac{du}{5\cos5x}$

Substitute: . $\int u^3\cos5x\left(\frac{du}{5\cos5x}\right) \;=\;\frac{1}{5}\int u^3\,du\;=\;\frac{1}{5}\cdot\frac{u^4}{4} + C$

Back-substitute: . $\frac{1}{20}\sin^45x + C$

$2. \int \tan^2x\,dx$

We have: . $\int\left(\sec^2x - 1\right)\,dx \;=\;\tan x - x + C$

$3. \int \frac{\sec^2x}{3+\tan x}dx$

Let $u = 3 + \tan x\quad\Rightarrow\quad du = \sec^2x\,dx\quad\Rightarrow\quad dx = \frac{du}{\sec^2x}$

Substitute: . $\int\frac{\sec^2x}{u}\left(\frac{du}{\sec^2x}\righ t) \;= \;\int\frac{du}{u} \;=\;\ln|u| + C$

Back-substitute: . $\ln|3 + \tan x| + C$

$4. \int \sin x \,\left[\cos x +\csc x \right]\,dx$

Multiply: . $\sin x\,\left[\cos x + \csc x\right] \;= \;(\sin x)(\cos x) + (\sin x)(\csc x) \;= \;\sin x \cos x + 1$

We have: . $\int\left(\sin x\cos x + 1\right)\,dx \;= \;\int\sin x\cos x\,dx + \int dx$

. . In the first integral, let $u = \sin x\quad\Rightarrow\quad du = \cos x\,dx$

. . Substitute: . $\int u\,du \;=\;\frac{1}{2}u^2 + c \;= \;\frac{1}{2}\sin^2x + C$

Answer: . $\frac{1}{2}\sin^2x + x + C$

5. Originally Posted by nirva

$3. \int \frac{\sec^2(x)}{3+\tan(x)}dx$
Let,
$u=3+\tan x$
Then,
$u'=\sec^2 x$
Thus, by substitution theorem,
$\int \frac{u'}{u}du=\int \frac{1}{u}du$
Which is,
$\ln |u|+C=\ln |3+\tan x|+C$