39For 3rd question refer to NCERT. It's a standard integral question..
For the first question use polynomial division.
1)\int \frac{x^3+x+1}{x^2-1}dx
2)\int \frac{tan\theta +tan^3\theta }{1+tan^3\theta }d\theta
3)\int \sqrt{cotx}dx
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6 Answers
39
493) let √cot x=t
cot x = t2
-cosec2xdx=2tdt
or, -(1+cot2x)dx=2tdt
or, dx=2tdt-(1+t4)
∫√cot x dx = - ∫ 2t2dt1+t4 = - ∫t2+11+t4dt - ∫t2-11+t4dt
∫t2+11+t4dt = ∫1+1t2t2+1t2dt (dividing by t2)
t2+1t2=(t-1t)2+2
so integral=∫1+1t2(t-1t)2+2dt
putting t-1t=u u will get the integral in the form of ∫duu2+2
similarly for second part of the integral divide by t2 and change the denominator as (t+1t)2-2
and u ll get same type of integral...
49arre no...
i din get any good substitution for the 3rd... so left with no hope i put t=√cot x :P
and LO! that solved the question :P
232nd:
put t = tanx
I =∫t1+t3 dt
=∫t(1+t)(t2-t+1) dt
use partial fractions now
11) ∫x3+x+1x2-1dx
Dividing N^{r} by D^{r} \int {x + \frac{2x+1}{x^{2} - 1}}dx = \int xdx + \int \frac{2x}{x^{2} - 1}dx + \int \frac{dx}{x^{2} - 1} = \frac{x^{2}}{2} + log\left|x^{2} - 1 \right| + \frac{1}{2}log\left|\frac{x - 1}{x + 1} \right| + C