Question

please help me with this integral problem

Answer 1

\[\int\limits_{?}^{?} \frac{ \tan^3x }{ \sqrt{secx} }\]

Answer 2

i am stuck at \[\int\limits_{?}^{?} \sec ^{3/2}x \sin ^{3}x dx\]

Answer 3

i dont know what to do after that

Answer 4

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

yes ?

Answer 5

multiply secx top and bottom so that substitution becomes easy :

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

Answer 6

Now substitute \(u =\sec x\)

Answer 7

you get
\[\int\dfrac{u^2-1}{u\sqrt{u}}du\]

which can be integrated easily

Answer 8

@ganeshie8 can you show me i want to make sure i am right?

Answer 9

i got \[\frac{ 2u^3 }{ 15 }-\frac{ 2 }{ 5 }u^^{5/2}\]

Answer 10

what?

Answer 11

\[\int\dfrac{u^2-1}{u\sqrt{u}}du = \int \dfrac{u^2-1}{u^{3/2}}du = \int u^{1/2} - u^{-3/2}du = \frac{u^{3/2 }}{3/2}-\frac{u^{-1/2}}{-1/2} + C\]

simplifying you get
\[\frac{2}{3}u^{3/2}+2u^{-1/2} + C\]

Answer 12

plug back \(u = \sec x\)

Answer 13

okay i was wrong lol thanks for the help

Answer 14

np :) let me knw if something is not clear

Answer 15

okay thank you