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(√cot x) is damn easy job :P
11 Answers
1write in parametric form! denominator becomes sin(x+aplha)
take x+alpha=t and write x=t-alpha.......expand numerator n integrate!
11. integrate
2.
FOR PRCTC
as alwayz pritish babu has done a nice work !!!!!!!!!!!!!!!!!
12) \int \left( sinxlogx-\frac{cosx}{x}\right)dx = \int sinx logx dx- \int \frac{cosx}{x}dx .......(1)
let I1 = \int sinx logx dx
considering sin x as 2nd and log x as first function we have BY USING BY PARTS,
I1 = logx\int sinx+\int \frac{cosx}{x}
now substituting the value of I1 in (1) we get,
\int \left( sinxlogx-\frac{cosx}{x}\right)dx =logx\int sinx+\int \frac{cosx}{x} - ∫cos xx = logx\int sinx
= - logx cosx
1LONG ONE THOUGH.
\int (\sqrt{tanx }+\sqrt{cotx}) dx = \int \frac{sinx+cosx}{\sqrt{sinxcosx}}dx now put sinx - cosx = t → (cosx + sinx)dx = dt
integral reduces to \sqrt{2}\int \frac{dt}{\sqrt{1-t^{2}}} = \sqrt{2}sin^{-1}t + c = \sqrt{2}sin^{-1}(sinx - cosx ) + c
similarly we find \int (\sqrt{tanx }-\sqrt{cotx})dx = -\sqrt{2}ln\left|sinx+cosx+\sqrt{sin2x} \right|+c
now \int \sqrt{cotx } = \frac{1}{2}\int 2\sqrt{cotx}dx = \frac{1}{2}\int (\sqrt{tanx}+\sqrt{cotx}) - ( \sqrt{tanx}-\sqrt{cotx})dx
subtituting the values of \int (\sqrt{tanx }+\sqrt{cotx}) dx and \int (\sqrt{tanx }-\sqrt{cotx})dx we get \int \sqrt{cotx } = \frac{1}{2}\left[ \sqrt{2}sin^{-1}(sinx-cosx)+\sqrt{2}ln\left|(sinx+cosx)+\sqrt{sin2x} \right|\right]+k,k = integ.constant
1whats the solution for 1st question ,
iam getting in the for of
K ∫√1- tan α. cos t dt
where t is x - α
