Question

Evaluate::$\int \frac{{\sin }^6\mathrm{x}+{\cos }^6\mathrm{x}}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}$

Answer 1

To evaluate: $\int \frac{{\sin }^6\mathrm{x}+{\cos }^6\mathrm{x}}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}$

$\mathrm{Let}\mathrm{I}$ $=$$\int \frac{{\sin }^6\mathrm{x}+{\cos }^6\mathrm{x}}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}$

$=$$\int \frac{{\left({\sin }^2\right)}^{3}x+{\left({\cos }^2\mathrm{x}\right)}^3}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}$

$=$$\int \frac{{\left({\sin }^2\right)}^{3}x+{\left({\cos }^2\mathrm{x}\right)}^3}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}$

$=$$\int \frac{\left({\sin }^2\right)\mathrm{x}+\left({\cos }^2\mathrm{x}\right)[{\left({\sin }^2\mathrm{x}\right)}^2+{\left({\cos }^2\mathrm{x}\right)}^2-{\sin }^2{\mathrm{xcos}}^2\mathrm{x}]}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}$ [Formula used: ${\mathbf{a}}^{\mathbf{3}}\mathbf{+}{\mathbf{b}}^{\mathbf{3}}\mathbf{=}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{(}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathit{a}\mathit{b}\mathbf{)}$]

$=$$\int \frac{[{\left(\mathrm{sinx}\right)}^4+{\left(\mathrm{cosx}\right)}^4-{\sin }^2{\mathrm{xcos}}^2\mathrm{x}]}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}$ [Identity used: $\mathbf{\left(}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{\right)}\mathbf{x}\mathbf{+}\mathbf{\left(}\mathbf{c}\mathbf{o}{\mathbf{s}}^{\mathbf{2}}\mathbf{\right)}\mathbf{x}\mathbf{=}\mathbf{1}$]

$=$ $\int \frac{[{\left(\mathrm{sinx}\right)}^4+{\left(\mathrm{cosx}\right)}^4+2{\sin }^2{\mathrm{xcos}}^2\mathrm{x}-3{\sin }^2{\mathrm{xcos}}^2\mathrm{x}]}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}$

$=$ $\int \frac{[({\mathrm{sinx}}^2+{\mathrm{cosx}}^2{)}^2-3{\sin }^2{\mathrm{xcos}}^2\mathrm{x}]}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}$

$=$$\int \frac{1-3{\sin }^2{\mathrm{xcos}}^2\mathrm{x}]}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}$ [Identity used: $\mathbf{\left(}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{\right)}\mathbf{x}\mathbf{+}\mathbf{\left(}\mathbf{c}\mathbf{o}{\mathbf{s}}^{\mathbf{2}}\mathbf{\right)}\mathbf{x}\mathbf{=}\mathbf{1}$]

$=$ $\int \frac{1}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}-\frac{3}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}$

$=$$\int \frac{{\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}-\frac{3}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}$ [Identity used: $\mathbf{\left(}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{\right)}\mathbf{x}\mathbf{+}\mathbf{\left(}\mathbf{c}\mathbf{o}{\mathbf{s}}^{\mathbf{2}}\mathbf{\right)}\mathbf{x}\mathbf{=}\mathbf{1}$]

$=$$\int \frac{{\sin }^2\mathrm{x}}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}+\frac{{\cos }^2\mathrm{x}}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}-\frac{3}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}$

$=$$\int \frac{{\sin }^2\mathrm{x}}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}+\int \frac{{\cos }^2\mathrm{x}}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}-\int \frac{3}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}$

$=$$\int \frac{1}{{\cos }^2\mathrm{x}}\mathrm{dx}+\int \frac{1}{{\sin }^2\mathrm{x}}\mathrm{dx}-\int \frac{3}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}$

$=$$\int {\sec }^2\mathrm{xdx}+\int {\mathrm{cosec}}^2\mathrm{xdx}-\int \frac{3}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}$ [Identity used: $\frac{1}{\mathrm{cosx}}=\mathrm{secx}$ and $\frac{1}{\mathrm{sinx}}=\cos ec\mathrm{x}$]

$=$$\int {\sec }^2\mathrm{xdx}+\int {\mathrm{cosec}}^2\mathrm{xdx}-\int \frac{3}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}$

$\mathrm{I}=\mathrm{tanx}-\mathrm{cotx}-3\mathrm{x}+\mathrm{C}$

Hence $\mathrm{I}=\mathrm{tanx}-\mathrm{cotx}-3\mathrm{x}+\mathrm{C}$