Question

I=tan^-1(sec3x+tan3x) dxpls ans this integration​

Answer 1

Question :

Integrate

$\sf \int \tan {}^{ - 1} ( \sec3x + \tan3x)dx$

Solution :

We have to integrate

$\sf \int \tan {}^{ - 1} ( \sec3x + \tan3x)dx$

•First solve the inverse trignometric part

$\sf \tan {}^{ - 1} ( \sec3x + \tan3x)$

$= \sf \tan {}^{ - 1} ( \dfrac{1}{ \cos3x} + \dfrac{ \sin3x}{ \cos3x} )$

$= \sf \tan {}^{ - 1} ( \dfrac{1 + \sin3x}{ \cos3x} )$

$= \sf \tan {}^{ - 1} ( \dfrac{ \cos {}^{2} \frac{3x}{2} + \sin {}^{2} \frac{3x}{2} + 2 \sin \frac{3x}{2} \cos \frac{3x}{2} }{ \cos {}^{2} \frac{3x}{2} - \sin {}^{2} \frac{3x}{2} } )$

$= \sf \tan {}^{ - 1} ( \dfrac{(cos \frac{3x}{2} + \sin \frac{3x}{2} ) {}^{2} }{ (cos \frac{3x}{2} + \sin \frac{3x}{2})(cos \frac{3x}{2} - \sin \frac{3x}{2})})$

$= \sf \tan {}^{ - 1} (\dfrac{\cos\frac{3x}{2} + \sin \frac{3x}{2}}{cos \frac{3x}{2} - \sin \frac{3x}{2}})$

$= \sf \tan {}^{ - 1} ( \dfrac{ 1 + \tan \frac{3x}{2} }{1 - \tan \frac{3x}{2} } )$

$= \sf \tan {}^{ - 1} (\dfrac{ \tan \frac{\pi}{4} + \tan \frac{3x}{2}}{1 - \tan \frac{\pi}{4} \tan \frac{3x}{2} })$

$= \sf \tan {}^{ - 1} ( \tan(\dfrac{\pi }{4} + \dfrac{3x}{2} )$

We know that

$\tan {}^{ - 1} ( \tan \: x) = x$

$= \sf \dfrac{\pi}{4} + \dfrac{3x}{2} ....(1)$

Now ,

•Put equation (1) value in integration:

$\implies \sf \int\tan {}^{ - 1} ( \sec3x + \tan3x)dx = \int( \dfrac{\pi}{4} + \dfrac{3x}{2})dx$

$= \int \frac{ \pi}{4} dx + \int \frac{3x}{2}dx$

$\sf = \dfrac{ \pi}{4}x + \dfrac{3x {}^{2} }{2} \times \dfrac{1}{2} + c$

$\sf = \dfrac{1}{4}(\pi \: x + 3x {}^{2} ) + c$

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Formula's used :

$1) \sf \: cos2x = \cos {}^{2} x - \sin {}^{2} x$

$2) \sf \sin2x = 2 \sin \: x \cos \: x$

$\sf3) \tan(x + y) = \dfrac{ \tan \: x + \tan \: y}{1 - \tan \: x \tan \: y}$

Answer 2

AnSwer:

†Our question is:

$\star \sf{ I = \int tan^{-1} ( sec(3x) + tan(3x) ) dx }$

†So we need to simplify,

$\star \sf{ ( sec(3x) + tan(3x) ) }$

$= \sf{ \dfrac{1}{cos(3x)} + \dfrac{sin(3x)}{cos(3x)} }$

$= \sf{ \dfrac{sin(3x) + 1}{cos(3x)}}$

†Now as we know,

$\star \sf{ \dfrac{1 + sin(2a)}{cos(2a)} = \dfrac{1 + tan(a)}{1 - tan(a)}}$

$\implies \sf{ \dfrac{1 + sin(3x)}{cos(3x)} = \dfrac{1 + tan \left(\frac{3x}{2}\right)}{1 - tan\left(\frac{3x}{2}\right)}}$

$\implies \sf{ \dfrac{1 + tan \left(\frac{3x}{2}\right)}{1 - tan\left(\frac{3x}{2}\right)} = tan\left( \frac{\pi}{4} + \frac{3x}{2} \right) }$

†Now we also know that,

$\star \sf{ tan^{-1} .tan(x) = x}$

Therefore,

$\sf{ = \int tan^{-1} tan\left( \frac{\pi}{4} + \frac{3x}{2} \right)\: dx }$

$\sf { = \int \left(\dfrac{\pi}{4} + \dfrac{3x}{2}\right) \: dx }$

$\sf { = \int \dfrac{\pi}{4} \: dx \: + \int \dfrac{3x}{2}\: dx }$

$\sf { = \dfrac{\pi}{4}(x) \: + \dfrac{3x^2}{4} + c }$

$\boxed{\sf{ I = \dfrac{x}{4} ( \pi + 3x ) + c }}$