# Some antiderivatives

1. Apr 1, 2008

### W3bbo

[SOLVED] Some antiderivatives

I've got a few antiderivatives to find, I've found most of them and they check out fine with my CAS, but three of them I'm having difficulties with.

The first:

1. The problem statement, all variables and given/known data

$$I = \int {{{\sec ^2 \left( x \right)} \over {\left( {1 + \tan \left( x \right)} \right)^3 }}} dx$$

3. The attempt at a solution

Using Integration by Substitution:

$$\displaylines{ I = \int {{{\sec ^2 \left( x \right)} \over {\left( {1 + \tan \left( x \right)} \right)^3 }}} dx \cr u = 1 + \tan \left( x \right) \cr {{du} \over {dx}} = \sec ^2 \left( x \right) \cr du = \sec ^2 \left( x \right)dx \cr I = \int {{{\sec ^2 \left( x \right)} \over {u^3 }}} dx \cr = \int {u^{ - 3} } du \cr = - {\textstyle{1 \over 2}}u^{ - 2} \cr = - {\textstyle{1 \over 2}}\left( {1 + \tan \left( x \right)} \right)^{ - 2} + C \cr = - {1 \over {2\left( {1 + \tan \left( x \right)} \right)^2 }} + C \cr}$$

Yet this does not coincide with the answer my CAS gets:

$${{ - {\mathop{\rm Cos}\nolimits} \left( {2x} \right) + {\mathop{\rm Sin}\nolimits} \left( {2x} \right)} \over {4 + 8{\mathop{\rm Cos}\nolimits} \left( x \right){\mathop{\rm Sin}\nolimits} \left( x \right)}}$$

Finally, for the other two problems I can't think of an approach. I've tried Integration by Parts on both of them, to no avail. I've also tried Substitution, but I get stuck quickly.

FWIW, they're:

$$I = \int {\sin \left( {2x} \right)\cos \left( x \right)} dx$$

And

$$I = \int {{\mathop{\rm Cos}\nolimits} ^2 \left( {2x} \right)} dx$$

Thanks!

Last edited: Apr 1, 2008
2. Apr 1, 2008

### sutupidmath

$$I = \int {{\mathop{\rm Cos}\nolimits} ^2 \left( {2x} \right)} dx$$

For this one you need to apply the double angle formula. That is

$$cos^{2}(x)=\frac{1+cos(2x)}{2}$$

$$I = \int {\sin \left( {2x} \right)\cos \left( x \right)} dx$$

For this one,

Hint: $$sin(2x)=2sin(x)cos(x)$$ than take a substitution of cos(x)=t, and you will be fine.

3. Apr 1, 2008

### Pere Callahan

Integration by part is a good way here. Just use it twice and observe that you get the original integral again. You can then algebraically solve for the integral.

For the second on you can either write it as cos(2x)cos(2x) and use the same trick, or you can try and employ some trigonometric indentities to get rid of the square, for example

$$\cos{2x}=\Re{e^{2ix}}=\Re{\left(\left[e^{ix}\right]^2\right)}=\Re{\left(\left[\cos x+i\sin x\right]^2\right)}=\Re{\left(\cos^2x-\sin^2x+2i\cos x\sin x\right)}=\cos^2x-\sin^2x=2\cos^2x-1$$

From this you see

$$\cos^2x=\frac{1}{2}(1+\cos{2x})$$
or
$$\cos^2{(2x)}=\frac{1}{2}(1+\cos{4x})$$

EDIT: Too slow

Last edited: Apr 1, 2008
4. Apr 1, 2008

### sutupidmath

I really dont want to act like a smart ass, like some of you guys are saying. However, here is my approach to it

$$I = \int {\sin \left( {2x} \right)\cos \left( x \right)} dx=2\int cos^{2}(x)sin(x)dx,u=cos(x)=>-du=sin(x)dx=>-2\int u^{2}du=-2\frac{u^{3}}{3}+C$$

5. Apr 1, 2008

### Pere Callahan

Smart ass!

No you're right, that is way easier than my suggestion.

6. Apr 2, 2008

### Gib Z

For number 1) Remember what the "+C" at the end actually means.

7. Apr 2, 2008

### W3bbo

Thanks all.

This does my working look alright?

$$\displaylines{ I = \int {\sin \left( {2x} \right)\cos \left( x \right)} dx \cr \sin \left( {2x} \right) = 2\sin \left( x \right)\cos \left( x \right) \cr I' = 2\sin \left( x \right)\cos ^2 \left( x \right) \cr u = \cos \left( x \right) \cr {{du} \over {dx}} = - \sin \left( x \right) \cr du = - \sin \left( x \right)dx \cr I = \int { - 2u} du \cr = - u^2 \cr = - \cos ^2 \left( x \right) \cr}$$

and

$$\displaylines{ I = \int {\cos ^2 \left( {2x} \right)} dx \cr \cos ^2 \left( x \right) = {\textstyle{1 \over 2}} + {\textstyle{1 \over 2}}\cos \left( {2x} \right) \cr I = \int {{\textstyle{1 \over 2}} + {\textstyle{1 \over 2}}cos\left( {4x} \right)} dx \cr = {\textstyle{1 \over 2}}\int {1 + \cos \left( {4x} \right)} dx \cr = {\textstyle{1 \over 2}}x + {\textstyle{1 \over 8}}\sin \left( {4x} \right) \cr}$$

8. Apr 2, 2008

### Gib Z

Well, for the sin(2x)cos x integral, you had it correct until the 3rd last line. Check that again.

The second one is correct, though both need the "+C" at the end. Well done.

9. Apr 2, 2008

### W3bbo

Ah, how careless of me.

And we're done:

$$\displaylines{ I = \int {\sin \left( {2x} \right)\cos \left( x \right)} dx \cr \sin \left( {2x} \right) = 2\sin \left( x \right)\cos \left( x \right) \cr I = \int {2\sin \left( x \right)\cos ^2 \left( x \right)dx} \cr u = \cos \left( x \right) \cr {{du} \over {dx}} = - \sin \left( x \right) \cr du = - \sin \left( x \right)dx \cr \cr I = \int { - 2u^2 } du \cr = - {\textstyle{2 \over 3}}u^3 + C \cr = - {\textstyle{2 \over 3}}\cos ^3 \left( x \right) + C \cr}$$

Thank you for all your help