Question

Please check the following integrtion

Answer 1

\[\tan ^{5}xdx \]
\[\tan(x)(\tan ^{2})^{2}dx\]
\[\tan(x)(\sec ^{2}(x)-1)\]
\[u=\sec ^{2}(x).....du=\tan(x)\]
\[\int\limits u ^{2}=1/3u ^{3}=\frac{ 1 }{ 3 }(\sec ^{2}(x)-1)^{3}+C\]

Answer 2

Third line should be sec^2(x)-1)^2

Answer 3

\[I=\int\limits \tan ^3x*\tan ^2xdx=\int\limits \tan ^3x \left( \sec ^2x-1 \right)dx=\int\limits \tan ^3x \sec ^2xdx-\int\limits \tan ^3xdx\]
\[=\frac{ \tan ^4x }{ 4 }-\int\limits \tan x \tan ^2x dx=\frac{ \tan ^4x }{ 4 }-\int\limits \tan x \left( \sec ^2x-1 \right)dx\]
\[=\frac{ \tan ^4x }{ 4 }-\frac{ \tan ^2x }{ 2 }+\int\limits \frac{ \sin x }{ \cos x }dx+c\]
\[=\frac{ \tan ^4x }{ 4 }-\frac{ \tan ^2x }{ 2 }-\ln \left| \cos x \right|+c\]

Answer 4

hmm seems you missed something

Answer 5

hmmmmm seems I missed alot

Answer 6

I would go with what suj did. separating it to \(\large tan^3x ~and~ tan^2x\)

Answer 7

Thanks