Integral of $\int \frac{x}{{sin}^{2} (x^{2})} d x$ $int x/sin^2(x^2)dx$ ?

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Answer 1 · 2018-06-27T16:13:11

$\int \frac{x}{{sin}^{2} (x^{2})} d x = - \frac{1}{2} cot (x^{2}) + c$ $int x/sin^2(x^2)dx = -1/2cot(x^2) + c$

Let $u = x^{2}$ $u = x^2$ , then $\frac{1}{2} u = x d x$ $1/2u = xdx$ :

$\frac{1}{2} \int \frac{1}{{sin}^{2} (u)} d u$ $1/2int 1/sin^2(u)du$

Substitute $\frac{1}{{sin}^{2} (u)} = {csc}^{2} (u)$ $1/sin^2(u) = csc^2(u)$ :

$\frac{1}{2} \int {csc}^{2} (u) d u$ $1/2int csc^2(u)du$

$- \frac{1}{2} cot (u) + c$ $-1/2cot(u) + c$

Reverse the substitution:

$\int \frac{x}{{sin}^{2} (x^{2})} d x = - \frac{1}{2} cot (x^{2}) + c$ $int x/sin^2(x^2)dx = -1/2cot(x^2) + c$

Integral of ∫xsin2(x2)dxint x/sin^2(x^2)dx?