We have to evaluate the following integral:
∫√sinx√sinx+√cosxdx
I tried this:
I multiplied both the numerator and denominator by secx
And substituted tanx=t.
But after that I got stuck.
The book where this is taken from gives the following as the answer: ln(1+t)−14ln(1+t4)+12√2lnt2−√2t+1t2+√2t+1−12tan−1t2+c where t=√cotx
Answer
I=∫√sinx√sinx+√cosxdx=∫√tanx1+√tanxdx
substitute tanx=t2 and dx=11+t4dt
I=∫t(1+t)(1+t4)dt=12∫((1+t4)+(1−t4))t(1+t)(1+t4)dt
=12∫t1+tdt+12∫(t−t2)(1+t2)1+t4dt
=12∫(1+t)−11+tdt+12∫t+t3−(t2−1)−t4−11+t4dt
=−t2+12ln|t+1|+14∫2t1+t4+12∫t31+t4dt−12∫t2−11+t4dt−12t+C
all integrals are easy except J=∫t2−11+t4dt=∫1−t−2(t+t−1)2−2dt=∫(t−t−1)′(t−t−1)2−2dt
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