1. Integral Multiplication

$\int tan\;x\sqrt{1+sec^4x}\;dx$

Thanks

2. I think you first need to get rid of that square root by doing some substitution.

Try $tan \theta = \sec^2 (x)$. It might not actually work out...but that 4th power has to be removed before you can replace that $\sqrt{1 + (\tan \theta)^2}$ with just $\sec \theta$.

At least that's what I would try. You just shouldn't use integration by parts if it looks like you can still substitute.

3. Originally Posted by TrevorP
I think you first need to get rid of that square root by doing some substitution.

Try $tan \theta = \sec^2 (x)$. It might not actually work out...but that 4th power has to be removed before you can replace that $\sqrt{1 + (\tan \theta)^2}$ with just $\sec \theta$.

At least that's what I would try. You just shouldn't use integration by parts if it looks like you can still substitute.
not sure that would make life any easier, after all, what would $\tan x$ become? $\sqrt{\tan \theta - 1}$. worst, what would $d \theta$ be? i never wrote anything down, but just thinking about it seems like a mess to me. unless you have some trick to simplify things

4. Originally Posted by polymerase
$\int tan\;x\sqrt{1+sec^4x}\;dx$
Okay, this will be a mess. It takes lots of substitutions, but having patience makes sense.

$\int {\sqrt {1 + \sec ^4 x} \tan x\,dx} = \int {\sqrt {1 + \cos ^4 x} \sec ^2 x\tan x\,dx} .$

Substitute $u=\tan x,$ the integral becomes

$\int {\frac{{u\sqrt {1 + \left( {u^2 + 1} \right)^2 } }}
{{u^2 + 1}}\,du} .$

Make another substitution according to $v=u^2+1,$ the integral becomes

$\frac{1}
{2}\int {\frac{{\sqrt {1 + v^2 } }}
{v}\,dv} .$

Now a final substitution defined by $\varphi ^2 = 1 + v^2.$ (We're gettin' close.)

$\frac{1}
{2}\int {\frac{{\varphi ^2 }}
{{\varphi ^2 - 1}}\,d\varphi } = \frac{1}
{2}\left[ {\int {d\varphi } + \frac{1}
{2}\left( {\int {\frac{1}
{{\varphi - 1}}\,d\varphi } - \int {\frac{1}
{{\varphi + 1}}\,d\varphi } } \right)} \right].$

Now this is easy, so we have

$\frac{1}
{2}\int {\frac{{\varphi ^2 }}
{{\varphi ^2 - 1}}\,d\varphi } = \frac{1}
{2}\left( {\varphi + \frac{1}
{2}\ln \left| {\frac{{\varphi - 1}}
{{\varphi + 1}}} \right|} \right) + k.$

After back substitute we finally-happily get that

$\int {\sqrt {1 + \sec ^4 x} \tan x\,dx} = \frac{1}
{2}\left( {\frac{{\sqrt {1 + \cos ^4 x} }}
{{\cos ^2 x}} + \frac{1}
{2}\ln \left| {\frac{{\sqrt {1 + \cos ^4 x} - \cos ^2 x}}
{{\sqrt {1 + \cos ^4 x} + \cos ^2 x}}} \right|} \right) \,+ \,k.$

5. $\tan \theta = \sec^2 x$
$\sec^2 \theta d \theta = 2 \sec x \sec x \tan x dx$
$\sec^2 \theta d \theta = 2 \tan \theta \tan x dx$
$\frac{\sec^2 \theta d \theta}{2 \tan \theta \tan x} = dx$

$\int \frac{\tan x \sqrt{1 + \tan^2 \theta} \sec^2 \theta d \theta }{2 \tan \theta \tan x} = \tfrac{1}{2} \int \frac{\sec^3 \theta d \theta}{\tan \theta }$

Now use integration by parts to solve that.

EDIT: Ok Krizalid already gave you his answer I guess I'll finish this.

$= \tfrac{1}{2} \int \underbrace{\sec^2 \theta}_{\tfrac{d}{d \theta} \tan \theta} \underbrace{\frac{\sec \theta}{\tan \theta }}_{\csc \theta} d \theta = \tfrac{1}{2} \Big( \tan \theta \csc \theta - \int \tan \theta (-) \csc \theta \cot \theta d \theta \Big)$
$= \tfrac{1}{2} \Big( \sec \theta + \int \csc \theta d \theta \Big) = \tfrac{1}{2} \Big( \sec \theta - ln|\csc \theta + \cot \theta| \Big)$
$= \tfrac{1}{2} \Big( \sec ( \arctan ( \sec^2 x ) ) - ln|\csc ( \arctan ( \sec^2 x ) ) + \cot ( \arctan ( \sec^2 x ) )| \Big)$

6. Originally Posted by Krizalid
Okay, this will be a mess. It takes lots of substitutions, but having patience makes sense.

$\int {\sqrt {1 + \sec ^4 x} \tan x\,dx} = \int {\sqrt {1 + \cos ^4 x} \sec ^2 x\tan x\,dx} .$

Substitute $u=\tan x,$ the integral becomes

$\int {\frac{{u\sqrt {1 + \left( {u^2 + 1} \right)^2 } }}
{{u^2 + 1}}\,du} .$

Make another substitution according to $v=u^2+1,$ the integral becomes

$\frac{1}
{2}\int {\frac{{\sqrt {1 + v^2 } }}
{v}\,dv} .$

Now a final substitution defined by $\varphi ^2 = 1 + v^2.$ (We're gettin' close.)

$\frac{1}
{2}\int {\frac{{\varphi ^2 }}
{{\varphi ^2 - 1}}\,d\varphi } = \frac{1}
{2}\left[ {\int {d\varphi } + \frac{1}
{2}\left( {\int {\frac{1}
{{\varphi - 1}}\,d\varphi } - \int {\frac{1}
{{\varphi + 1}}\,d\varphi } } \right)} \right].$

Now this is easy, so we have

$\frac{1}
{2}\int {\frac{{\varphi ^2 }}
{{\varphi ^2 - 1}}\,d\varphi } = \frac{1}
{2}\left( {\varphi + \frac{1}
{2}\ln \left| {\frac{{\varphi - 1}}
{{\varphi + 1}}} \right|} \right) + k.$

After back substitute we finally-happily get that

$\int {\sqrt {1 + \sec ^4 x} \tan x\,dx} = \frac{1}
{2}\left( {\frac{{\sqrt {1 + \cos ^4 x} }}
{{\cos ^2 x}} + \frac{1}
{2}\ln \left| {\frac{{\sqrt {1 + \cos ^4 x} - \cos ^2 x}}
{{\sqrt {1 + \cos ^4 x} + \cos ^2 x}}} \right|} \right) + k.$
nice.

could you have dealt with something like $\int \frac {\sqrt{1 + u^4}}u~du$ in the same way? because we could have gotten that by one substitution, $u = \sec x$

7. Originally Posted by Jhevon
could you have dealt with something like $\int \frac {\sqrt{1 + u^4}}u~du$ in the same way? because we could have gotten that by one substitution, $u = \sec x$
Sure, just substitute $v^2=1+u^4\implies v\,dv=2u^3\,du,$ then

$\int {\frac{{\sqrt {1 + u^4 } }}
{u}\,du} = \frac{1}
{2}\int {\frac{{v^2 }}
{{v^2 - 1}}\,dv} ,$
and the rest follows.

(Of course you meant to say $u^2=\tan x.$ )