Intergration by parts question from A2 core 4

[Sink41]

I need to find the integral of

(x^2) 2 (secx)^2 tanx dx

said aloud: x squared times by two times by sec squared x times by tan x

I tried to use the [tex] function but failed misserably, hope you understand what i mean from what I've written above.

 

[Saketh]

In TeX:
$$\int 2 x^2 \sec^2{x} \tan{x} \,dx$$
You may have to carry out integration by parts several times before you get an answer.

 

[courtrigrad]

$$2 \int x^{2} \sec^{2}{x} \tan{x} \,dx$$. Let $$u = x^{2}, du = 2x dx, dv = \sec^{2}x \tan x, v = \frac{1}{2}\sec^{2} x$$.

So we have $$\int udv = \frac{x^{2}\sec^{2}x}{2} - \int {x\sec^{2}x} = \frac{x^{2}\sec^{2}x}{2} - x\tan x - \ln| \cos x | + C$$

$$\int 2x^{2} \sec^{2}{x} \tan{x} \,dx = x^{2}\sec^{2}x - 2x\tan x- 2\ln| \cos x | + C$$

 

[Sink41]

$$2 \int x^{2} \sec^{2}{x} \tan{x} \,dx$$. Let $$u = x^{2}, du = 2x dx, dv = \sec^{2}x \tan x, v = \frac{1}{2}\sec^{2} x$$.

So we have $$\int udv = \frac{x^{2}\sec^{2}x}{2} - \int {x\sec^{2}x} = \frac{x^{2}\sec^{2}x}{2} - x\tan x - \ln| \cos x | + C$$

$$\int 2x^{2} \sec^{2}{x} \tan{x} \,dx = x^{2}\sec^{2}x - 2x\tan x- 2\ln| \cos x | + C$$

courtrigrad thanks for your help. I realized i had made a mistake when intergrating the

$$\sec^{2}{x} \tan{x}$$

bit and differentiated some of it by mistake... anyway got into a big mess. But when reading through your answer realized what i did wrong. btw, not trying to be pedantic, but you made a small mistake when intergrating tanx its

$$\int 2x^{2} \sec^{2}{x} \tan{x} \,dx = x^{2}\sec^{2}x - 2x\tan x- 2\ln| \sec x | + C$$

 

[courtrigrad]

$$\int \tan x dx = \int \frac{\sin x}{\cos x} dx, u = \cos x, du = -\sin x$$. $$-\int \frac{du}{u} = -\ln|u| = -\ln|\cos x|$$. So its correct.

 

[Sink41]

ah yes forgot -ln|cosx| = ln|secx| sorry.