Question

Please, help
\[\int\dfrac{1}{3sinx+1}dx\]

Answer 1

Faiq: Students generally must learn how to solve problems themselves. You might suggest their use of www.integral-calculator.com only after your student has made and shared some effort to integrate on his own.

Answer 2

Thoughts:
Have you tried integration by parts?
Any chance that you could simplify the integration by multiplying numerator and denominator of \[\frac{ 1 }{ 3\sin x + 1 }\]

Answer 3

hint: multiply and divide by 3sinx-1

Answer 4

by (3 xin x - 1)?

Answer 5

Great minds think alike.

Answer 6

^yes :)

Answer 7

:)

Answer 8

Oh maybe we can do this
\[\large \rm \int \frac{1}{3sinx+1} *1\]

Answer 9

\[\large \rm \int\limits \frac{1}{3sinx+1} *1\] is incomplete (and not proper) unless you include that differential, "dx."

Answer 10

\[\large \rm \int\limits \frac{1}{3sinx+1} *1dx\]

Answer 11

but wouldn't we get a much more complex integral in its solution by parts?

Answer 12

by doing this \(\large \rm \int\limits \frac{1}{3sinx+1} *1dx\) we'll encounter this-\[\int\limits_{}^{}\frac{ 3cosx }{ (3sinx+1)^2 }dx\] which will get complex

Answer 13

it must be this*\(\int\limits_{}^{}\frac{x (3cosx )}{ (3sinx+1)^2 }dx\)

Answer 14

no that is alright^^
but when you solve it further using integration by parts then you'll encounter this^

Answer 15

Unfortunately, we go nowhere there.

Answer 16

(3sinx +1) (3sinx -1) = 9sin^2 x -1

Answer 17

then??

Answer 18

If we multiple by 3sin x +1, the integrand becomes
\(\dfrac{3sinx +1}{9sin^2x +6sin x+1}\) , then??

Answer 19

No outcome!!