MHB 8.2.5.evaluate int 64x sec^2(4x) dx

[karush]

Evaluate $I=\displaystyle \int 64x\sec^2(4x) \, dx$

ok well first $64 \displaystyle\int x\sec^2(4x) \, dx$

off hand not sure what trig id to use or if we need to

 

[skeeter]

integration by parts …

$u = 16x \implies du = 16 \, dx$
$dv = 4\sec^2(4x) \, dx \implies v = \tan(4x)$

$\displaystyle \int 16x \cdot 4\sec^2(4x) \, dx = 16x \cdot \tan(4x) - \int 16\tan(4x) \, dx$

can you finish up?

 

[karush]

integration by parts …

$u = 16x \implies du = 16 \, dx$
$dv = 4\sec^2(4x) \, dx \implies v = \tan(4x)$

$\displaystyle \int 16x \cdot 4\sec^2(4x) \, dx = 16x \cdot \tan(4x) - \int 16\tan(4x) \, dx$

can you finish up?

so if ... (from plug in table)
$\displaystyle 16 \int \tan \left(4x\right)dx
=16\left[-\dfrac{1}{4}\ln \left|\cos \left(4x\right)\right|\right]+C
=-4\ln \left|\cos \left(4x\right)\right|+C$
Hence
$I=16x \cdot \tan(4x)+4\ln \left|\cos \left(4x\right)\right|+C$

Spoiler: w|A returned this

anyway not sure...

 

[skeeter]

not sure about what?

 

[karush]

isn't W|A answer different or is it??

 

[skeeter]

W/A omitted the absolute value around the log argument.
Why it did, I don't know.