Mathematics

# $\int {\dfrac{{{{\sin }^6}x + {{\cos }^6}x}}{{{{\sin }^2}x.{{\cos }^2}x}}dx}$

##### SOLUTION

$\int{\dfrac{{{\sin }^{6}}x+{{\cos }^{6}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}}$

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

$\because {{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$

${{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$

$\because {{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}} \right)}^{3}}={{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{3}}-3{{\sin }^{2}}x{{\cos }^{2}}x\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)$

$={{1}^{3}}-3\sin x\cos x\left( 1 \right)$

$=1-3\sin x\cos x$

Therefore,

$\int{\dfrac{{{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}} \right)}^{3}}}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{\left( \dfrac{1-3{{\sin }^{2}}x{{\cos }^{2}}x}{si{{n}^{2}}x{{\cos }^{2}}x} \right)}dx$

$=\int{\left( \dfrac{1}{{{\sin }^{2}}x{{\cos }^{2}}x}-\dfrac{3{{\sin }^{2}}x{{\cos }^{2}}x}{si{{n}^{2}}x{{\cos }^{2}}x} \right)}dx$

$=\int{\left( \dfrac{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}{si{{n}^{2}}x{{\cos }^{2}}x}-3 \right)}dx$

$=\int{\left( \dfrac{{{\sin }^{2}}x}{si{{n}^{2}}x{{\cos }^{2}}x}+\dfrac{{{\cos }^{2}}x}{si{{n}^{2}}x{{\cos }^{2}}x}-3 \right)dx}$

$=\int{\left( \dfrac{1}{{{\cos }^{2}}x}+\dfrac{1}{{{\sin }^{2}}x}-3 \right)dx}$

$=\int{\left( {{\sec }^{2}}x+\cos e{{c}^{2}}x-3 \right)}dx$

$=\int{\sec ^{2}xdx+\int{\cos e{{c}^{2}}xdx}}-\int{3dx}$

$=\tan x-\cot x-3x+c$

Hence, the value of integral is $\tan x-\cot x-3x+c$.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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