Question

integrate: (0 to pi/4) 1/cos^3x (whole root)(2sin 2x):

Answer 1

Integral ( 0 to pi / 4 ) 1 / $cos^{3}x$$\sqrt{2sin2x}$ is equal to 12 / 5$\sqrt{2}$

Consider the integral without the limits.

We know 1 / $cosx$ =  $secx$

• ∫ $\frac{dx}{cos^{3}(x)\sqrt{2sin2x} }$  =  ∫ sec^3x dx/( $\sqrt{2sin2x}$) =

• ∫ sec ^ 3x dx / (  $\sqrt{2sin2x}$ )  =  ∫ sec ^ 3x dx / ( $\sqrt{2sinxcosx}$)

Multipying and dividing by $secx$ ,

• ∫ sec ^ 3x dx / ( $\sqrt{2sinxcosx}$)  =  ∫ sec ^ 3x . sec x dx /  sec x . $\sqrt{2sinxcosx}$

• ∫ sec ^ 3x . sec x dx / sec x . $\sqrt{2sinxcosx}$   = ∫ sec ^ 3x . sec x dx / ($\sqrt{2\frac{sinx}{cosx}\frac{cosx}{cosx} }$)

• ∫ sec ^ 3x . sec x dx / ($\sqrt{2\frac{sinx}{cosx}\frac{cosx}{cosx} }$)   =  ∫ sec ^ 2x . sec ^ 2x dx / $\sqrt{2tanx}$

Also, sec ^ 2x  =  1 + tan ^ x

• ∫ sec ^ 2x . sec ^ 2x dx / $\sqrt{2tanx}$  = ∫sec ^ 2x ( 1 +  tan ^ x ) dx / $\sqrt{2tanx}$

Substituting t  =  tanx ,  dt  =  sec ^ 2 ( x ) dx

• ∫$sec^{2}x$ ( 1 + tan ^ x ) dx / $\sqrt{2tanx}$  =   $\frac{1}{\sqrt{2} } (\frac{2}{5} t^{\frac{5}{2} } x + 2\sqrt{t} )$

Substituting back t  =  tanx , gives

• ∫$sec^{2}x$(  1 +  tan ^ x ) dx / $\sqrt{2tanx}$   =  $\frac{1}{\sqrt{2} } (\frac{2}{5} tan^{\frac{5}{2} } x + 2\sqrt{tanx} )$

Applying limits x from 0 to $\frac{\pi }{4}$ , tan$\frac{\pi }{4}$ = 1 and tan 0 = 0,

Integral Value  =  1/$\sqrt{2}$ ( 2 / 5 + 2 )  =  12 / 5$\sqrt{2}$