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(cotx/sinx-tan^2x- tanx/cos+2x/cos^2x)dx

Guest Aug 29, 2017

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 #1
avatar+90988 
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(cotx/sinx-tan^2x- tanx/cos+2x/cos^2x)dx

NOTE I have added an x after cos as I think that you forgot it.  :)

 

\(\int (\frac{cotx}{sinx}-tan^2x-\frac{ tanx}{cosx}+\frac{2x}{cos^2x})dx\\ =\int \frac{cotx}{sinx}dx-\int tan^2x dx-\int\frac{ tanx}{cosx} dx+\int\frac{2x}{cos^2x}dx\\ =\int \frac{cotx}{sinx}dx-\int tan^2x dx-\int\frac{ tanx}{cosx} dx+\int\frac{2x}{cos^2x}dx\\~\\ -----\\ \int \frac{cotx}{sinx}dx=\int \frac{cosx}{sin^2x}dx=\int cosx(sinx)^{-2}dx=-(sinx)^{-1}+c_1=-cosecx+c_1\\ \int tan^2x dx=\int \frac{sin^2x}{cos^2x} dx=\int\;(sec^2x-1)dx=tanx-x+c_2\\ \int\frac{ tanx}{cosx} dx=\int\frac{ sinx}{cos^2x} dx=(cosx)^{-1}+c_3=secx+c_3\\~\\ \text{So far I have }\\ =\:-cosecx-(tanx-x)-(secx)+\int\frac{2x}{cos^2x}dx\\ =\:-cosecx-tanx+x-secx+\int\frac{2x}{cos^2x}dx \)

 

\(=\:-cosecx-tanx+x-secx+\int\frac{2x}{cos^2x}dx\\ consider\;\;\;\int\frac{2x}{cos^2x}dx\\ =\int(2x(sec^2x)dx\\ \text{Solve using integration by parts}\\ u=2x \qquad v'=sec^2x\\ u'=2 \qquad v=tanx\\ \int\;uv'dx=uv-\int(vu')dx\\ =2xtanx-\int(2tanx)dx\\ =2xtanx-2\int(\frac{sinx}{cosx})dx\\ =2xtanx+2ln(cosx))+c_4\\~\\ \text{So my final answer is}\\ =\:-cosecx-tanx+x-secx+2xtanx+2ln(cosx)+c\\ \)

Melody  Aug 29, 2017
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 #1
avatar+90988 
+3
Best Answer

(cotx/sinx-tan^2x- tanx/cos+2x/cos^2x)dx

NOTE I have added an x after cos as I think that you forgot it.  :)

 

\(\int (\frac{cotx}{sinx}-tan^2x-\frac{ tanx}{cosx}+\frac{2x}{cos^2x})dx\\ =\int \frac{cotx}{sinx}dx-\int tan^2x dx-\int\frac{ tanx}{cosx} dx+\int\frac{2x}{cos^2x}dx\\ =\int \frac{cotx}{sinx}dx-\int tan^2x dx-\int\frac{ tanx}{cosx} dx+\int\frac{2x}{cos^2x}dx\\~\\ -----\\ \int \frac{cotx}{sinx}dx=\int \frac{cosx}{sin^2x}dx=\int cosx(sinx)^{-2}dx=-(sinx)^{-1}+c_1=-cosecx+c_1\\ \int tan^2x dx=\int \frac{sin^2x}{cos^2x} dx=\int\;(sec^2x-1)dx=tanx-x+c_2\\ \int\frac{ tanx}{cosx} dx=\int\frac{ sinx}{cos^2x} dx=(cosx)^{-1}+c_3=secx+c_3\\~\\ \text{So far I have }\\ =\:-cosecx-(tanx-x)-(secx)+\int\frac{2x}{cos^2x}dx\\ =\:-cosecx-tanx+x-secx+\int\frac{2x}{cos^2x}dx \)

 

\(=\:-cosecx-tanx+x-secx+\int\frac{2x}{cos^2x}dx\\ consider\;\;\;\int\frac{2x}{cos^2x}dx\\ =\int(2x(sec^2x)dx\\ \text{Solve using integration by parts}\\ u=2x \qquad v'=sec^2x\\ u'=2 \qquad v=tanx\\ \int\;uv'dx=uv-\int(vu')dx\\ =2xtanx-\int(2tanx)dx\\ =2xtanx-2\int(\frac{sinx}{cosx})dx\\ =2xtanx+2ln(cosx))+c_4\\~\\ \text{So my final answer is}\\ =\:-cosecx-tanx+x-secx+2xtanx+2ln(cosx)+c\\ \)

Melody  Aug 29, 2017

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