11GOOD WORK DONE BY SIR
BUT I WOULD LIKE TO COMMENT
PUT x=tanθ
dx=sec2θ
then u would have to integrate
∫cotθcos4θdθ
do it by parts[1][1][1]
5 Answers
62\int \frac{dx}{x(x^2+1)^3}
= \int \frac{xdx}{x^2(x^2+1)^3}
x^2=t
2xdx=dt
= 1/2 \int \frac{dt}{t(t+1)^3}
Now it can be done by partial fractions?
62\frac{1}{t(t+1)^3}=\frac{t+1-t}{t(t+1)^3}
\\=\frac{1}{t(t+1)^2}-\frac{1}{(t+1)^3} \\=\frac{t+1-t}{t(t+1)^2}-\frac{1}{(t+1)^3} \\=\frac{1}{t(t+1)}-\frac{1}{(t+1)^2}-\frac{1}{(t+1)^3} \\=\frac{t+1-t}{t(t+1)}-\frac{1}{(t+1)^2}-\frac{1}{(t+1)^3} \\=\frac{1}{t}-\frac{1}{(t+1)}-\frac{1}{(t+1)^2}-\frac{1}{(t+1)^3}
1162mani i thnk there wud be cos 4 theta term...
but ofcourse not to say that your method is wrong :)
11yes sir u r right i m8issed that
ur method is the best 4 this 1 [1]
