Question

Why did I get this wrong?

Answer 1

\[\int\limits_{}^{}3\tan^5(x)dx\]

Answer 2

I get \[\frac{ 3\sec^4(x)}{ 4 }-3\sec^2x+\ln|sexc|+C\]

Answer 3

So this is what I did:

\[3\int\limits_{}^{}(\tan^2x)^2tanx dx\]

Answer 4

so I could use the pythagorean identity

Answer 5

Show your working.

Answer 6

\[3\int\limits_{}^{}(1-\sec^2x)^2tanxdx\]

Answer 7

\[3\int\limits_{}^{}(\sec^4x-2\sec^2x+1)tanxdx\]

Answer 8

\[3\int\limits_{}^{}(\sec^4x tanx-2\sec^2x tanx+tanx)dx\]

Answer 9

\[3\int\limits_{}^{}\sec^4x tanx-\int\limits_{}^{}2\sec^2x tanx+\int\limits_{}^{} tanxdx\]

Answer 10

for the first integral: U substitution
u=secx
du=secx tanx
\[3\int\limits_{}^{}u^3du\]
= \[\frac{ 3u^4 }{ 4}= \frac{ 3\sec^4x }{ 4 }\]

Answer 11

second integral:

\[u= tanx\]
\[du= \sec^2x du\]

=

\[3u^2\]

Answer 12

third integral:
\[\ln|secx|\]

Answer 13

Looks fine. But keep in mind the original 3 outside the integral has to multiply by every term in the result.

Answer 14

ohhh I forgot the 3 in front of the ln >.<