Помогите решить интегрирование тригонометрической функции, пример: [tex]\int\limits {\frac{dx}{sin^3xcos^5x} }[/tex] . Ответ должен получится: [tex]-\frac{1}{2tg^2x} +3ln|tgx|+\frac{3}{2} tg^2x+\frac{1}{4} tg^4x+C[/tex]. Created by milakamilat1m. matematika-ru

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NNNLLL54

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Ответ:

$\displaystyle \int \frac{dx}{sin^3x\cdot cos^5x}=\int \frac{\frac{dx}{cos^8x}}{\frac{sin^3x\cdot cos^5x}{cos^8x}}=\int \frac{(\frac{1}{cos^2x})^3\cdot \frac{dx}{cos^2x}}{tg^3x}=\int \frac{(1+tg^2x)^3\cdot d(tgx)}{tg^3x}=\\\\\\=\Big[\ t=tgx\ \Big]=\int \frac{(1+t^2)^3}{t^3}\, dt=\int \frac{1+3t^2+3t^4+t^6}{t^3}\, dt=\\\\\\=\int \Big(t^{-3}+\frac{3}{t}+3t+t^3\Big)\, dt=\frac{t^{-2}}{-2}+3\, ln|\, t\, |+\frac{3t^2}{2}+\frac{t^4}{4}+C=$

$=-\dfrac{1}{2\, tg^2x}+3\, ln|tgx|+\dfrac{3}{2}\, tg^2x+\dfrac{1}{4}\, tg^4x+C$

2 votes Thanks 1

LymarIvan

Ответ:

Пошаговое объяснение:

$\int {\frac{dx}{\sin^3x\cos^5x} } \, =\int {\frac{1}{\sin^3x\cos^3x }* \frac{dx}{\cos^2x} } \, =\int {\frac{d(\rm tg \; x)}{\sin^3x\cos^3x }} \, =\int {\frac{d(\rm tg \; x)}{\frac{\sin^3x}{\cos^3x}*(cos^3x\cos^3x) } } \,=\\\int {\frac{d(\rm tg \; x)}{\rm tg^3x(cos^2x)^3} } \, =\int {\frac{d(\rm tg \; x)}{\rm tg^3x} } *(\frac{1}{\cos^2x})^3 \, =\int {\frac{(1+\rm tg^2x)^3}{\rm tg^3x} } \, d(\rm tg \; x)=\begin{Vmatrix} \rm tg \; x=t\end{Vmatrix}=$

$\int {\frac{(1+t^2)^3}{t^3} } \, dt=\int {\frac{1+3t^2+3t^4+t^6}{t^3} } \, dt=\int {(\frac{1}{t^3}+\frac{3}{t} +3t+t^3) } \, dt =\int {(t^{-3}+\frac{3}{t} +3t+t^3) } \, dt=\\=-\frac{1}{2}t^{-2}+3\ln|t| +\frac{3t^2}{2}+\frac{t^3}{4} +C=-\frac{1}{2\rm tg^2x}+3\ln|\rm tg \; x|+\frac{3}{2}\rm tg^2x+\frac{1}{4}\rm tg^4x+C$

2 votes Thanks 1

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