Mathematics

# Solve $\displaystyle\int {\dfrac{{6{{\sin }^3}x + 5{{\cos }^3}x}}{{{{\sin }^2}x\,{{\cos }^2}x}}\,\,dx}$

##### SOLUTION

Given that:

$=\int {\dfrac{{6{{\sin }^3}x + 5{{\cos }^3}x}}{{{{\sin }^2}x\,{{\cos }^2}x}}\,\,dx}$

$=\int{\dfrac{6\,{\sin}x}{{\cos}^2{x}}}\,dx +\,\int{\dfrac{5\,{\cos}x}{{\sin}^2{x}}}\,dx$

$=\int6{\tan{x}\sec{x}}\, dx +\int5{\cot{x}\csc{x}}\,dx$

$\left[\because \int{\tan{x}\sec{x}}\,=\sec{x}, \int{\cot{x}\csc{x}}\,dx = -\csc{x}\right]$

$=6\sec{x}\,+5(\,-\csc{x})\, +constant$

$=6\sec{x}-5\csc{x}\,+constant$

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

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