I do not know where I went wrong?

Question

I was doing a question in which one step included solving this integral

$$ \int \frac{t}{\cos^2 t}\,dt $$

So what book did was $$ \int t\sec^2 t\,dt $$

And then applied integration by parts .

Whereas what I did was $$ \int \frac{t} {1-\sin^2t}\, dt $$

$$ \int\frac{t}{(1+ \sin t)(1- \sin t)}\, dt $$

And then using partial fraction . $$ \frac{A}{1+ \sin t} + \frac{B}{1-\sin t} = t $$

$A = \frac{3\pi}{4}$ and $B = \frac{\pi}{4}$

Which when put in $$ \frac{A}{1+ \sin t} + \frac{B}{1-\sin t} = t $$ it did not satisfy the equation.

Which I think means that the values of $A$ and $B$ are incorrect.

So could someone tell where I went wrong?

Answer 1

There are no constants $A$ and $B$ such that $\frac A {1+\sin t} +\frac B {1-\sin t}=t$ so your method does not work.

Answer 2

For any real number $x\neq\pm1$, you have$$\frac1{1-x^2}=\frac12\left(\frac1{1+x}+\frac1{1-x}\right)$$and therefore$$\frac1{1-\sin^2t}=\frac{\frac12}{1+\sin t}+\frac{\frac12}{1-\sin t}.$$So,$$\frac t{1-\sin^2 t}=\frac{\frac t2}{1+\sin t}+\frac{\frac t2}{1-\sin t}.$$

Answer 3

Alternative solution by parts:

Set $u(t)=t,v'(t)=\frac{1}{\cos^2t}.$

Then $u'(t)=1, v(t)=\tan t=\frac{\sin t}{\cos t}$ and

$$\int \frac{t}{\cos^2t}dt=t\tan t-\int\frac{\sin t}{\cos t}dt=t\tan t+\log(\cos t)+c$$