Question

integral of (1/(x^2*cos^2(x))+3sin(2x)) dx

Answer 1

\[\int \dfrac{1}{x^2cos^2x}+3sin(2x) dx\] ??

Answer 2

yes

Answer 3

You don't have elementary result for this.

Answer 4

\[\int\limits \frac{ 1 }{ x ^{2}\cos ^{2} }dx+3\times \int\limits \sin(2x)dx\]

Answer 5

http://www.wolframalpha.com/input/?i=int+%281%2Fx^2cos^2x%29%2B3sin%282x%29dx

Answer 6

\[\int\limits_{1/(x^2\cos^2(x)) dx}^{?}\]

Answer 7

this takes too much time to solve &write here
u should use rule:
\[\int\limits udv=uv-\int\limits vdu\]

Answer 8

yes i know how to solve the second part of sin , but the first part we should use integration by parts but it doesnt work i dont know how to solve the first part !

Answer 9

i told u how to solve the first one
the rule that i wrote
u should use that
1/x^2=u
1/cos^2dx=dv

Answer 10

the second one is easy
the first one take too much time
let me write it myself & also u tried yourself

Answer 11

wait a moment plz

Answer 12

u should write (1/x^2)*(1+tan^2)
do u understand why?

Answer 13

then with the rule i told u:
\[u=\frac{ 1 }{ x ^{2} }\]
\[dv=1+\tan ^{2}dx\]

Answer 14

no i didn t understand

Answer 15

there is a rule
the rule was what i write to u
\[\int\limits u \times dv=u \times v -\int\limits v \times du\]
its a rule

Answer 16

from this rule u should find what is U&dv in your question(remember i m solving the first part and this rule is for the firs part)

Answer 17

(U) is 1/x^2
(dv) is (1+tan^2)dx
OK?

Answer 18

just u should find (du) and (v)
then just rewrite in the rule

Answer 19

\[u=\frac{ 1 }{ x ^{2} }\rightarrow du= -\frac{ 1 }{ x }\]
\[dv=(1+\tan ^{2})dx \rightarrow v=\int\limits 1+\tan ^{2}dx=tanx\]
got it?

Answer 20

now just rewrite in the rule(the frame i told u)

Answer 21

do u understand?

Answer 22

yes thank you

Answer 23

yourwelcome