Question

integrate

Answer 1

\[\int\limits_{}^{} \sqrt{tanx}dx\]

Answer 2

http://math2.org/math/integrals/more/tan.htm hopefully this helps

Answer 3

It is an extremely messy integral.

Answer 4

let \[tanx = t^2\]
\[\sec^2 x dx=2t dt\]
so
\[ dx=\frac{2t dt}{\sec^2 x}\]
\[ dx=\frac{2t dt}{1+t^2}\]

Answer 5

sorry ,
\[dx=\frac{2tdt}{1+t^4}\]

Answer 6

\[=\int\limits_{}^{}\frac{t* 2tdt}{1+t^4}\]
\[=\int\limits_{}^{}\frac{2t^2dt}{1+t^4}\]
\[=\int\limits_{}^{}\frac{(t^2+1+t^2-1)dt}{1+t^4}\]
\[=\int\limits_{}^{}\frac{(t^2+1)dt}{t^4+1}+\int\limits_{}^{}\frac{(t^2-1)dt}{t^4+1}\]

Answer 7

dividing by t ^2
\[=\int\limits_{}^{}\frac{(1+\frac{1}{t^2})dt}{t^2+\frac{1}{t^2}}+\int\limits_{}^{}\frac{(1-\frac{1}{t^2})dt}{t^2+\frac{1}{t^2}}\]
\[=\int\limits_{}^{}\frac{(1+\frac{1}{t^2})dt}{t^2+\frac{1}{t^2}-2+2}+\int\limits_{}^{}\frac{(1-\frac{1}{t^2})dt}{t^2+\frac{1}{t^2}+2-2}\]
\[=\int\limits_{}^{}\frac{(1+\frac{1}{t^2})dt}{(t-\frac{1}{t})^2+2}+\int\limits_{}^{}\frac{(1-\frac{1}{t^2})dt}{(t+\frac{1}{t})^2-2}\]
now can you integrate ?

Answer 8

for first part substitute :
\[u=t-\frac{1}{t}\]
for second part substitute :
\[v=t+\frac{1}{t}\]