Question

Integral 2sin2x-cosx/6-cos^2x-4sinx

Answer 1

Refer to attachment.....

Answer 2

Answer:

$2 \ln \left(\sin ^{2} \theta-4 \sin \theta+5\right)+7 \tan { }^{-1}(\sin \theta-2)$

Step-by-step explanation:

Solution

$\int \frac{2 \sin 2 \theta-\cos \theta}{6-\cos ^{2} \theta-4 \sin \theta} d \theta \int \frac{2 \sin 2 \theta-\cos \theta}{-4 \sin \theta-\cos ^{2} \theta+6} d \theta$

$=-\int \frac{2 \sin 2 \theta-\cos \theta}{4 \sin \theta+\cos ^{2} \theta-6} d \theta$

Substituting

$\sin 2 \theta=2 \sin \theta \cos \theta \cos ^{2} \theta=1-\sin ^{2} \theta$

$=\int \cos \theta\left(-\frac{4 \sin \theta-1}{\sin ^{2} \theta-4 \sin \theta+5}\right) d \theta$

Substituting

$\mathrm{u}=\sin \theta \rightarrow \mathrm{d} \theta=\frac{1}{\cos \theta} \mathrm{du}$

$=\left[-\int \frac{4 \mathrm{u}-1}{\mathrm{u}^{2}-4 \mathrm{u}+5} \mathrm{du}\right]$

$4 \mathrm{u}-1 \text { as } 2(2 \mathrm{u}-\mathrm{u})+7$

$=\int\left(\frac{2(2 \mathrm{u}-4)}{\mathrm{u}^{2}-4 \mathrm{u}+5}+\frac{7}{\mathrm{u}^{2}-4 \mathrm{u}+5}\right) \mathrm{du}$

$=4 \int \frac{\mathrm{u}-2}{\mathrm{u}^{2}-4 \mathrm{u}+5} \mathrm{du}+7 \int \frac{1}{\mathrm{u}^{2}-4 \mathrm{u}+5} \mathrm{du}$

$\mathrm{v}=\mathrm{u}^{2}-4 \mathrm{u}+5 \rightarrow \mathrm{du}=\frac{1}{2 \mathrm{u}-4} \mathrm{dv}$

$=4 \int \frac{1}{2 \mathrm{v}} \mathrm{dv}$

$=\frac{1}{2} \int \frac{1}{\mathrm{v}} \mathrm{dv}$

$=\frac{\ln \mathrm{v}}{2}$

$=\ln \frac{\left(\mathrm{u}^{2}-4 \mathrm{u}+5\right)}{2}$

$=\int \frac{1}{(\mathrm{u}-2)^{2}+1} \mathrm{du} \mathrm{v}=\mathrm{u}-2 \rightarrow \mathrm{du}=\mathrm{dv}$

$=\int \frac{1}{\mathrm{v}^{2}+1} \mathrm{dv}$

$=\tan { }^{-1}(\mathrm{v}) \mathrm{v}=\mathrm{u}-2$

$=\tan { }^{-1}(\mathrm{u}-2)$

$=4 \ln \frac{\left(\mathrm{u}^{2}-4 \mathrm{u}+5\right)}{2}+7 \tan ^{-1}(\mathrm{u}-2)$

$=-2 \ln \left(\mathrm{u}^{2}-4 \mathrm{u}+5\right)-7 \tan ^{-1}(\mathrm{u}-2)$

$=2 \ln \left(\sin ^{2} \theta-4 \sin \theta+5\right)+7 \tan { }^{-1}(\sin \theta-2)$

#SPJ2