Question

Evaluate the indefinite integral.

1/cos^5(x)dx

Answer 1

sounds like a calc II problem to me

Answer 2

yeah its a calc 2 problem....its giving me trouble.

Answer 3

\[\int\limits_{}^{}\sec^5(x) dx=\int\limits_{}^{}\sec^4(x) \sec(x) dx=\int\limits_{}^{}(1+\tan^2(x))^2\sec(x) dx\]
\[=\int\limits_{}^{}(1+2\tan^2(x)+\tan^4(x))\sec(x) dx\]
\[=\int\limits_{}^{}\sec(x) dx+2 \int\limits_{}^{}\tan^2(x) \sec(x) dx+\int\limits_{}^{}\tan^4(x) \sec(x) dx\]
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first lets do
\[\int\limits_{}^{}\sec(x) dx=\int\limits_{}^{}\sec(x) \cdot \frac{\sec(x)+\tan(x)}{\sec(x)+\tan(x)} dx\]
let \[u=\sec(x)+\tan(x) => du=(\tan(x)\sec(x)+\sec^2(x) )dx=\sec(x)(\tan(x)+\sec(x))dx\]
so we have
\[\int\limits_{}^{}\sec(x) dx=\int\limits_{}^{}\sec(x) \cdot \frac{\sec(x)+\tan(x)}{\sec(x)+\tan(x)} dx\]
\[=\int\limits_{}^{}\frac{du}{u}=\ln|u|+c_1=\ln|\sec(x)+\tan(x)|+c_1\]
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Answer 4

now we look at
\[2\int\limits_{}^{}\tan^2(x) \sec(x) dx\]

Answer 5

reminds me of why i hate this subject so very much

Answer 6

let's look at \[\int\limits_{}^{}\tan^2(x)\sec(x) dx\]

Answer 7

and then when we find antiderivative we will multiply by 2

Answer 8

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

Answer 9

so you have to integral by parts the sec^2 and sec^3

Answer 10

but we know \[\int\limits_{}^{}\sec^3(x) dx-\int\limits_{}^{}\sec(x) dx=\int\limits_{}^{}\sec^3(x) dx-\ln|\sec(x)+\tan(x)| +c_2\]
lol i love my constants
anyways now we to do
\[\int\limits_{}^{}\sec^3(x) dx=\int\limits_{}^{}\sec^2(x)\sec(x) dx\]
we need integration by parts here

Answer 11

integral by parts?

Answer 12

i just use the distributive property

Answer 13

lol i mean integrate by part

Answer 14

\[\int\limits_{}^{}\sec^2(x) \sec(x) dx=\tan(x) \sec(x)-\int\limits_{}^{}\tan(x) \sec(x) \tan(x) dx\]
\[=\tan(x) \sec(x)-\int\limits_{}^{}\tan^2(x) \sec(x) dx\]
\[=\tan(x)\sec(x)-\int\limits_{}^{}(\sec^2(x)-1)\sec(x) dx\]
\[=\tan(x)\sec(x)-\int\limits_{}^{}\sec^3(x) dx+\int\limits_{}^{}\sec(x) dx\]
\[=\tan(x)\sec(x)-\int\limits_{}^{}\sec^3(x) dx+\ln|\sec(x)+\tan(x)|\]
add \[\int\limits_{}^{}\sec^3(x) dx\] on both sides then divide by 2 to get
\[\int\limits_{}^{}\sec^3(x) dx=\frac{1}{2}\tan(x)\sec(x)+\frac{1}{2}\ln|\sec(x)+\tan(x)|\]

Answer 15

\[\int\limits\limits_{}^{}\tan^2(x)\sec(x) dx=\int\limits\limits_{}^{}(\sec^2(x)-1)\sec(x) dx=\int\limits\limits_{}^{}\sec^3(x) dx-\int\limits\limits_{}^{}\sec(x) dx \]
\[=\frac{1}{2}\tan(x)\sec(x)+\frac{1}{2}\ln|\sec(x)+\tan(x)|-\ln|\sec(x)+\tan(x)|+c_3\]

so anyways don't forget we actually had
\[2\int\limits_{}^{}\tan^2(x)\sec(x) dx=2 \cdot (\frac{1}{2} \tan(x)\sec(x)+\frac{1}{2}\ln|\sec(x)+\tan(x)|-\ln|\sec(x)+\tan(x)|)+c_3\]

Answer 16

\[2\int\limits_{}^{}\tan^2(x)\sec(x) dx=\tan(x)\sec(x)-\ln|\sec(x)+\tan(x)|+c_3\]

Answer 17

ok then the last part is
\[\int\limits_{}^{}\tan^4(x) \sec(x) dx\]
i leave this to you as an exercise

Answer 18

okay thanks you for showing me the steps on how i can reach this problem and ones similar to it