Integrating $\cos 2\theta$ and $\tan\theta$

[juantheron]

$(1)\;\; \displaystyle \int \cos 2\theta\cdot \ln \left(\frac{\cos \theta +\sin \theta}{\cos \theta -\sin \theta}\right)d\theta$

$(2)\;\; \displaystyle \int \frac{\tan 2\theta}{\sqrt{\sin^6 \theta +\cos ^6 \theta}}d\theta$

I have Tried for (II) :: $\displaystyle \int \frac{\tan 2\theta}{\sqrt{\sin^6 \theta +\cos ^6 \theta}}d\theta$

We can write $\displaystyle \sqrt{\sin^6 \theta +\cos ^6 \theta} = 1-3\sin^2 \theta .\cos^2 \theta = 1-\frac{3}{4}\sin^2 2\theta = \frac{4-3\sin^2 \theta}{4}$

$\displaystyle \int\frac{\sin 2\theta}{\cos 2\theta}\cdot \frac{2}{\sqrt{3-4\sin^2 2\theta}}d\theta $Now I am struch here,

Similarly My Try for (I) one

$\displaystyle \int \cos 2\theta\cdot \ln \left(\frac{\cos \theta +\sin \theta}{\cos \theta -\sin \theta}\right)d\theta = \int \cos2 \theta \cdot \ln\left(\frac{1+\tan \theta}{1-\tan \theta}\right)d\theta$

Now How can i solve after that

Help me

Thanks

 

[MarkFL]

For the first one, I would try integration by parts, where:

$$u=\ln\left(\frac{\cos(\theta)+\sin(\theta)}{ \cos(\theta)-\sin(\theta)} \right)=\ln\left(\frac{1+\sin(2\theta)}{ \cos(2\theta)} \right)\,\therefore\,du= \frac{\sin(2\theta)-1}{\sin(2\theta)}$$

$$dv=\cos(2\theta)\,d\theta\,\therefore\,v=\frac{1}{2}\sin(2\theta)$$

 

[Prove It]

$(1)\;\; \displaystyle \int \cos 2\theta\cdot \ln \left(\frac{\cos \theta +\sin \theta}{\cos \theta -\sin \theta}\right)d\theta$

$(2)\;\; \displaystyle \int \frac{\tan 2\theta}{\sqrt{\sin^6 \theta +\cos ^6 \theta}}d\theta$

I have Tried for (II) :: $\displaystyle \int \frac{\tan 2\theta}{\sqrt{\sin^6 \theta +\cos ^6 \theta}}d\theta$

We can write $\displaystyle \sqrt{\sin^6 \theta +\cos ^6 \theta} = 1-3\sin^2 \theta .\cos^2 \theta = 1-\frac{3}{4}\sin^2 2\theta = \frac{4-3\sin^2 \theta}{4}$

$\displaystyle \int\frac{\sin 2\theta}{\cos 2\theta}\cdot \frac{2}{\sqrt{3-4\sin^2 2\theta}}d\theta $Now I am struch here,

Similarly My Try for (I) one

$\displaystyle \int \cos 2\theta\cdot \ln \left(\frac{\cos \theta +\sin \theta}{\cos \theta -\sin \theta}\right)d\theta = \int \cos2 \theta \cdot \ln\left(\frac{1+\tan \theta}{1-\tan \theta}\right)d\theta$

Now How can i solve after that

Help me

Thanks

No need for integration by parts (at least not until after a great deal of simplification)...

\displaystyle \begin{align*} \frac{\cos{(\theta)} + \sin{(\theta)} }{ \cos{(\theta)} - \sin{(\theta)} } &= \frac{\left[ \cos{(\theta)} + \sin{(\theta)} \right] ^2}{ \left[ \cos{(\theta)} - \sin{(\theta)} \right] \left[ \cos{(\theta)} + \sin{(\theta)} \right] } \\ &= \frac{\cos^2{(\theta)} + 2\sin{(\theta)}\cos{(\theta)} + \sin^2{(\theta)}}{\cos^2{(\theta)} - \sin^2{(\theta)}} \\ &= \frac{1 + \sin{(2\theta)}}{\cos{(2\theta)}} \\ &= \frac{1 + \sin{(2\theta)}}{\sqrt{1 - \sin^2{(2\theta)}}} \end{align*}

So for

\displaystyle \begin{align*} \int{ \cos{(2\theta)} \ln{ \left[ \frac{\cos{(\theta)} + \sin{(\theta)}}{\cos{(\theta)} - \sin{(\theta)}} \right] } \, d\theta} = \frac{1}{2} \int{ 2\cos{(2\theta)}\ln{ \left[ \frac{1 + \sin{(2\theta)}}{\sqrt{1 - \sin^2{(2\theta)}}} \right] }\,d\theta} \end{align*}

Let \displaystyle \begin{align*} x = \sin{(2\theta)} \implies dx = 2\cos{(2\theta)} \, d\theta \end{align*} and the integral becomes

\displaystyle \begin{align*} \frac{1}{2} \int{ \ln{ \left[ \frac{1 + x}{\sqrt{1 - x^2}} \right] }\,dx } &= \frac{1}{2} \int{ \ln{ \left( 1 + x \right) } - \ln{ \left[ \left( 1 - x^2 \right) ^{\frac{1}{2}} \right] } \, dx } \\ &= \frac{1}{2} \int{ \ln{ \left( 1 + x \right) } - \frac{1}{2} \ln{ \left[ \left( 1 - x \right) \left( 1 + x \right) \right] } \, dx } \\ &= \int{ \ln{ \left( 1 + x \right) } - \frac{1}{2}\ln{ \left( 1 - x \right) } - \frac{1}{2} \ln{ \left( 1 + x \right) } \,dx} \\ &= \frac{1}{2} \int{ \frac{1}{2} \ln{ \left( 1 + x \right) } - \frac{1}{2} \ln{ \left( 1 - x \right) } \, dx } \\ &= \frac{1}{4} \int{ \ln{ \left( 1 + x \right) } - \ln{ \left( 1 - x \right) } \, dx} \end{align*}

Both of these integrals should be very easy to evaluate...

 

[Prove It]

$(2)\;\; \displaystyle \int \frac{\tan 2\theta}{\sqrt{\sin^6 \theta +\cos ^6 \theta}}d\theta$

I have Tried for (II) :: $\displaystyle \int \frac{\tan 2\theta}{\sqrt{\sin^6 \theta +\cos ^6 \theta}}d\theta$

We can write $\displaystyle \sqrt{\sin^6 \theta +\cos ^6 \theta} = 1-3\sin^2 \theta .\cos^2 \theta = 1-\frac{3}{4}\sin^2 2\theta = \frac{4-3\sin^2 \theta}{4}$

$\displaystyle \int\frac{\sin 2\theta}{\cos 2\theta}\cdot \frac{2}{\sqrt{3-4\sin^2 2\theta}}d\theta $Now I am struch here,

Your approach is a good one, assuming your identities are correct (I haven't checked). From here

\displaystyle \begin{align*} \int{ \frac{2\sin{(2\theta)}}{\cos{(2\theta)}\sqrt{3 - 4\sin^2{(2\theta)}}}\,d\theta} &= -\int{\frac{-2\sin{(2\theta)}}{\cos{(2\theta)}\sqrt{3 - 4 \left[ 1 - \cos^2{(2\theta)} \right] }}\,d\theta} \\ &= -\int{\frac{-2\sin{(2\theta)}}{\cos{(2\theta)}\sqrt{3 - 4 + \cos^2{(2\theta)}}}\,d\theta} \\ &= -\int{\frac{-2\sin{(2\theta)}}{\cos{(2\theta)}\sqrt{\cos^2{( 2 \theta )} - 1}}\,d\theta} \end{align*}

Now a substitution of the form \displaystyle \begin{align*} \sec{(x)} = \cos{(2\theta)} \implies \sec{(x)}\tan{(x)}\,dx &= -2\sin{(2\theta)} \end{align*} is appropriate, giving

\displaystyle \begin{align*} -\int{\frac{-2\sin{(2\theta)}}{\cos{(2\theta)}\sqrt{\cos^2{(2 \theta )} - 1}}\,d\theta} &= -\int{\frac{\sec{(x)}\tan{(x)}}{\sec{(x)}\sqrt{\sec^2{(x)} - 1}}\,dx} \\ &= -\int{\frac{\tan{(x)}}{\sqrt{\tan^2{(x)}}}\,dx} \\ &= -\int{\frac{\tan{(x)}}{\tan{(x)}}\,dx} \\ &= -\int{1\,dx} \\ &= -x + C \\ &= -\textrm{arsec}\,{\left[ \cos{(2\theta)} \right] } + C \end{align*}