6∫ (x2 + 20) dx(xsinx + 5cosx)2
ab clear hai??!!!
:P
11 Answers
23u hav written the same thing , and still it aint clear
write it using fractions eg: xy
62\frac{x^2+20}{(x\sin x+5\cos x)^2}=\frac{\frac{x^2+20}{x^2+25}}{\frac{(x\sin x+5\cos x)^2}{x^2+25}}=\frac{\frac{x^2+20}{x^2+25}}{sin^2(x+tan^{-1}(5/x))}
Now substitute \\t=x+tan^{-1}(5/x)\\\Rightarrow \frac{dt}{dx}=1+\frac{1}{1+25/x^2}\times(-5/x^2)=1+\frac{-5}{x^2+25}=\frac{x^2+20}{x^2+25}
Hence the given integral becomes
\int cosec^2t dt=-cot(t)+c
[1]
49x2+20(xsinx+5cosx)2=x2+25(xsinx+5cosx)2 - 5(xsinx+5cosx)2
x2+25=x2(sin2x+cos2x)+25(sin2x+cos2x) + 10xsinxcosx - 10xsinxcosx = (xsinx+5cosx)2+(xcosx-5sinx)2
so frst part is x2+25(xsinx+5cosx)2=(xsinx+5cosx)2+(xcosx-5sinx)2(xsinx+5cosx)2
which can be identified as ddx[5sinx-xcosxxsinx+5cosx]
but i am stuck with the part 5(xsinx+5cosx)2
btw" great soln by nishant bhaiya! isn't it??
1this can also be proceed by multiplying and dividing X^8cos(x) in numerator and denominator