-0-tan x-sec x-cos xdx-

∫_0^πtan x/sec x+cos xdx=∫_0^πsin x/cos x/1/cos x+cos xdx =∫_0^πsin x/1+cos^2xdx Put cos x = t Then sin x dx = dt x = 0 ⇒ t = cos 0 = 1 ∫_0^πsin x/1+cos^2xd ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Property 1

∫0πtan x/sec ...

Question

$\int_{0}^{π} \frac{t a n x}{s e c x + c o s x} d x =$

\frac{π}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 π

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{π}{2}$
$\int_{0}^{π} \frac{t a n x}{s e c x + c o s x} d x = \int_{0}^{π} \frac{\frac{s i n x}{c o s x}}{\frac{1}{c o s x} + c o s x} d x = \int_{0}^{π} \frac{s i n x}{1 + c o s^{2} x} d x$
Put cos x = t
Then - sin x dx = dt
x = 0 $\Rightarrow$ t = cos 0 = 1
$\int_{0}^{π} \frac{s i n x}{1 + c o s^{2} x} d x = \int_{1}^{- 1} \frac{1}{1 + t^{2}} (- d t) = [t a n^{- 1} (t)]_{- 1}^{1} = t a n^{- 1} (1) - t a n^{- 1} (- 1)$
$= t a n^{- 1} 1 + t a n^{- 1} 1 = 2 t a n^{- 1} (1) = 2. \frac{π}{4} = \frac{π}{2}$

Suggest Corrections

Similar questions

\int \frac{tan x}{sec x} cos x d x

Q. Using properties of integrals, calculate

\int 3 e^{x} + 5 cos x - 10 {sec}^{2} x d x

If $A = \int_{0}^{π} \frac{s i n x}{s i n x + c o s x} d x$

$B = \int_{0}^{π} \frac{s i n x}{s i n x - c o s x} d x$

$\frac{π}{4} \int 0 l o g (\frac{s i n x + c o s x}{c o s x}) d x$

\int \frac{cos \sqrt{x}}{\sqrt{x}} d x =

Join BYJU'S Learning Program

∫π0tan xsec x+cos xdx=

The correct option is C π2∫π0tan xsec x+cos xdx=∫π0sin xcos x1cos x+cos xdx=∫π0sin x1+cos2xdx Put cos x = t Then - sin x dx = dt x = 0 ⇒ t = cos 0 = 1 ∫π0sin x1+cos2xdx=∫−11 11+t2(−dt)=[tan−1(t)]1−1=tan−1(1)−tan−1(−1) =tan−11+tan−11=2 tan−1(1)=2.π4=π2

$\int_{0}^{π} \frac{t a n x}{s e c x + c o s x} d x =$