substitution rule

**jsu03** · December 9th 2008, 07:37 PM

Can someone check if these are correct. If they're not, please help me how to solve it the right way. Thanks a lot!!!
1. integral e^cost sint dt
u=cost d/du=-sint du=-sintdt , -du=sint dt
=integral e^u -du= -integral e^cost + c
2. integral sin(ln x)/x dx
u=ln x d/du=1/x du=1/xdx, -1/xdu=dx
=-1/x integral u du = -1/x(lnx)+c
3. integral (1 + tan thelta)^5 sec^2 thelta d thelta
u= 1+ tan thelta, d/du=-ln(cosx) du=-ln(cosx)dx, -du=ln(cos)xdx
= integral u^5 (-du) = -1/6(1 + tan thelta)^6 +C

**Prove It** · December 9th 2008, 07:55 PM

Originally Posted by jsu03

Can someone check if these are correct. If they're not, please help me how to solve it the right way. Thanks a lot!!!
1. integral e^cost sint dt
u=cost d/du=-sint du=-sintdt , -du=sint dt
=integral e^u -du= -integral e^cost + c
2. integral sin(ln x)/x dx
u=ln x d/du=1/x du=1/xdx, -1/xdu=dx
=-1/x integral u du = -1/x(lnx)+c
3. integral (1 + tan thelta)^5 sec^2 thelta d thelta
u= 1+ tan thelta, d/du=-ln(cosx) du=-ln(cosx)dx, -du=ln(cos)xdx
= integral u^5 (-du) = -1/6(1 + tan thelta)^6 +C

It's so much easier to read if you use LaTeX...

1. $\int{e^{\cos{t}}\sin{t}\,dt}$

Let $u = \cos{t}$ so $\frac{du}{dt} = -\sin{t}$

$\int{e^{\cos{t}}\sin{t}\,dt} = -\int{-e^{\cos{t}}\sin{t}\,dt}$

$= -\int{e^u \frac{du}{dt}\, dt}$

$= -\int{e^u \, du}$

$= -e^u + C$

$= -e^{\cos{t}} + C$ .

**Prove It** · December 9th 2008, 07:58 PM

Originally Posted by jsu03

Can someone check if these are correct. If they're not, please help me how to solve it the right way. Thanks a lot!!!
1. integral e^cost sint dt
u=cost d/du=-sint du=-sintdt , -du=sint dt
=integral e^u -du= -integral e^cost + c
2. integral sin(ln x)/x dx
u=ln x d/du=1/x du=1/xdx, -1/xdu=dx
=-1/x integral u du = -1/x(lnx)+c
3. integral (1 + tan thelta)^5 sec^2 thelta d thelta
u= 1+ tan thelta, d/du=-ln(cosx) du=-ln(cosx)dx, -du=ln(cos)xdx
= integral u^5 (-du) = -1/6(1 + tan thelta)^6 +C

2. $\int{\frac{\sin{\ln{x}}}{x}\, dx}$

Let $u = \ln{x}$ so $\frac{du}{dx} = \frac{1}{x}$ .

So $\int{\frac{\sin{\ln{x}}}{x}\, dx} = \int{\sin{u} \frac{du}{dx}\, dx}$

$= \int{\sin{u} \, du}$

$= -\cos{u} + C$

$= -\cos{\ln{x}} + C$ .

**Prove It** · December 9th 2008, 08:03 PM

Originally Posted by jsu03

Can someone check if these are correct. If they're not, please help me how to solve it the right way. Thanks a lot!!!
1. integral e^cost sint dt
u=cost d/du=-sint du=-sintdt , -du=sint dt
=integral e^u -du= -integral e^cost + c
2. integral sin(ln x)/x dx
u=ln x d/du=1/x du=1/xdx, -1/xdu=dx
=-1/x integral u du = -1/x(lnx)+c
3. integral (1 + tan thelta)^5 sec^2 thelta d thelta
u= 1+ tan thelta, d/du=-ln(cosx) du=-ln(cosx)dx, -du=ln(cos)xdx
= integral u^5 (-du) = -1/6(1 + tan thelta)^6 +C

3. $\int{(1 + \tan{\theta})^5 \sec^2{\theta}\, d\theta}$

Let $u = 1 + \tan{\theta}$ so $\frac{du}{d\theta} = \sec^2{\theta}$ .

So $\int{(1 + \tan{\theta})^5 \sec^2{\theta}\, d\theta} = \int{u^5 \frac{du}{d\theta}\, d\theta}$

$= \int{u^5 \, d\theta}$

$= \frac{1}{6}u^6 + C$

$= \frac{1}{6}(1 + \tan{\theta})^6 + C$ .

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