Question

Solve the following$\int_0^ \frac{\pi}{2} sin2x \: logtanx \: dx$

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle\int_0^\dfrac{\pi}{2} \rm \: sin2x \:log tanx \: dx$

Let assume that

$\rm \:I \: = \: \displaystyle\int_0^\dfrac{\pi}{2} \rm \: sin2x \:log tanx \: dx - - - (1)$

We know,

$\boxed{\tt{ \displaystyle\int_0^a\rm f(x) \: dx \: = \: \displaystyle\int_0^a\rm f(a - x) \: dx \: }}$

So, using this property of definite integrals, we get

$\rm \:I \: = \: \displaystyle\int_0^\dfrac{\pi}{2} \rm \: sin2\bigg(\dfrac{\pi}{2} - x \bigg) \:log tan\bigg(\dfrac{\pi}{2} - x \bigg) \: dx$

$\rm \: I = \displaystyle\int_0^\dfrac{\pi}{2} \rm sin(\pi - 2x) \: logcotx \: dx$

$\rm \: I = \displaystyle\int_0^\dfrac{\pi}{2} \rm sin2x \: logcotx \: dx - - - - (2)$

On adding equation (1) and (2), we get

$\rm \: 2I = \displaystyle\int_0^\dfrac{\pi}{2} \rm sin2x \: logtanx \: dx + \displaystyle\int_0^\dfrac{\pi}{2} \rm sin2x \: logcotx \: dx$

$\rm \: 2I = \displaystyle\int_0^\dfrac{\pi}{2} \rm sin2x \:( logtanx + logcotx) \: dx$

We know,

$\boxed{\tt{ \: \: logx + logy = log(x + y) \: \: }}$

So, using this, we get

$\rm \: 2I = \displaystyle\int_0^\dfrac{\pi}{2} \rm sin2x \:log(tanx \times cotx) \: dx$

$\rm \: 2I = \displaystyle\int_0^\dfrac{\pi}{2} \rm sin2x \:log\bigg(tanx \times \frac{1}{tanx} \bigg) \: dx$

$\rm \: 2I = \displaystyle\int_0^\dfrac{\pi}{2} \rm sin2x \:log\bigg(1\bigg) \: dx$

$\rm \: 2I = 0$

$\rm\implies \:I = 0$

Hence,

$\rm\implies \: \: \boxed{\tt{ \: \rm \: \displaystyle\int_0^\dfrac{\pi}{2} \rm \: sin2x \:log tanx \: dx = 0 \: \: }}$

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ADDITIONAL INFORMATION

$\rm \: \displaystyle\int_a^b\rm f(x) \: dx \: = \: \displaystyle\int_a^b\rm f(y) \: dy \:$

$\rm \: \displaystyle\int_a^b\rm f(x) \: dx \: = \: - \: \displaystyle\int_b^a\rm f(x) \: dx \:$

$\rm \: \displaystyle\int_a^b\rm f(x) \: dx \: = \:\: \displaystyle\int_a^b\rm f(a + b - x) \: dx \:$

$\rm \: \displaystyle\int_{ - a}^a\rm f(x) \: dx \: = \:\: \displaystyle\int_0^a\rm f( - x) \: dx \: \: if \: f( - x) = f(x) \:$

$\rm \: \displaystyle\int_{ - a}^a\rm f(x) \: dx \: = \:\:0 \: dx \: \: if \: f( - x) = - f(x) \:$

$\rm \: \displaystyle\int_{0}^{2a}\rm f(x) \: dx \: = \:\:0 \: dx \: \: if \: f(2a - x) = - f(x) \:$

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Answer:

What is integral tan x / (1 + m^2 tan^2 x) dx from 0 to pi/2?

I=∫π20tanx1+m2tan2xdx

Convert Everything into sinx and cosx

I=∫π20sinxcosx1+m2sin2xcos2xdx

I=∫π20sinxcosxcos2x+m2sin2xdx

Now simply take sin2x=t

That would also imply cos2x=1−t

Taking Derivative of Both Sides…

[mat...

Let I=∫π20tan(x)1+m2tan2(x)dx

=∫π20sin(x)cos(x)1+m2sin2(x)cos2(x)dx

=∫π20sin(x)cos(x)cos2(x)+m2sin2(x)dx

=∫π20sin(x)cos(x)cos2(x)+m2(1−cos2(x))dx

=∫π20sin(x)cos(x)m2+(1−m2)cos2(x)dx

Let cos(x)=y

⟹−sin(x)dx=dy

At x=0 , y=1

and x=π2,y=0

Substituting above value in I , we get,

I=∫01−ym2+(1−m2)y2dy

=∫102(1−m2)y2(1−m2)(m2+(1−m2)y2)dy

=∫1012(1−m2)(m2+(1−m2)y2)d(m2+(1−m2)y2)

=12(1−m2)ln(m2+(1−m2)y2)∣∣∣10

=12(1−m2)ln(m2+(1−m2))−12(1−m2)ln(m2)

=−11−m2ln(m)

⟹I=1m2−1ln(m)

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Step-by-step explanation:

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