If a= displaystyle ∫ 01(c o s (sin x)/sec â¡ x)dx, then the value of a2+(c o s)2(sin 1) is equal to

Question

If a= displaystyle ∫ 01(c o s (sin x)/sec â¡ x)dx, then the value of a2+(c o s)2(sin 1) is equal to (A) 0 (B) 1 (C) sin(1) (D) sin(sin 1)

Tardigrade Admin · Accepted Answer

a =∫01 cos ( sin x) cos x d x Let sin x= t ⇒ cos x d x = dt ∴ a =∫0 sin 1 cos t dt =( sin t)0 sin 1 = sin ( sin 1)- sin 0 = sin ( sin 1) ∴ a 2+ cos 2( sin 1)= sin 2 θ+ cos 2 θ=1 ;(θ= sin 1)

Q. If $a = \int_{0}^{1} \frac{cos ( s in x )}{sec x} d x,$ then the value of $a^{2} + (cos)^{2} (s in 1)$ is equal to

$0$

$1$

$s in (1)$

$s in (s in 1)$

$a = \int_{0}^{1} cos (sin x) cos x d x$
Let $sin x = t$
$\Rightarrow cos x d x = d t$
$∴ a = \int_{0}^{s i n 1} cos t d t = (sin t)_{0}^{s i n 1}$
$= sin (sin 1) - sin 0$
$= sin (sin 1)$
$∴ a^{2} + cos^{2} (sin 1) = sin^{2} θ + cos^{2} θ = 1; (θ = sin 1)$

Q. If a=∫01​sec⁡xcos(sinx)​dx, then the value of a2+(cos)2(sin1) is equal to

0

1

sin(1)

sin(sin1)

a=∫01​cos(sinx)cosxdx Let sinx=t ⇒cosxdx=dt ∴a=∫0sin1​costdt=(sint)0sin1​ =sin(sin1)−sin0 =sin(sin1) ∴a2+cos2(sin1)=sin2θ+cos2θ=1;(θ=sin1)

Q. If $a = \int_{0}^{1} \frac{cos ( s in x )}{sec x} d x,$ then the value of $a^{2} + (cos)^{2} (s in 1)$ is equal to

$0$

$1$

$s in (1)$

$s in (s in 1)$

$a = \int_{0}^{1} cos (sin x) cos x d x$
Let $sin x = t$
$\Rightarrow cos x d x = d t$
$∴ a = \int_{0}^{s i n 1} cos t d t = (sin t)_{0}^{s i n 1}$
$= sin (sin 1) - sin 0$
$= sin (sin 1)$
$∴ a^{2} + cos^{2} (sin 1) = sin^{2} θ + cos^{2} θ = 1; (θ = sin 1)$