Question

S sec(ax +b) tan(ax + b) dx​

Answer 1

Answer:

ANSWER

∫sin(ax+b)cos(ax+b)dx

u=sin(ax+b)⇒dx=

acos(ax+b)

1

du

∫sin(ax+b)cos(ax+b)=

a

1

∫udu

=

2a

u

2

=

2a

sin

2

(ax+b)

∫sin(ax+b)cos(ax+b)dx=

2a

sin

2

(ax+b)

+C

Step-by-step explanation:

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Answer 2

Answer: The value of integration is (sec(ax+b))/a + c, where c is constant of integration

Step-by-step explanation:

$\int{sec(ax+b)tan(ax+b)} \, dx$

⇒ $\int{\frac{1}{cos(ax+b)}*\frac{sin(ax+b)}{cos(ax+b)} } \, dx$

⇒ $\int{\frac{sin(ax+b)}{cos^{2} (ax+b)} } \, dx$

Let ax + b = u

Differentiating both sides with respect to x,

⇒ $a = \frac{du}{dx}$

⇒ $dx = \frac{du}{a}$

Substituting values of ax + b and dx in terms of u and du :

⇒ $\int{\frac{sinu}{cos^{2}u} } \, \frac{du}{a}$

⇒ $\int{\frac{sinu}{acos^{2}u} } \,du$

Let cos(u) = t

Differentiating both sides with respect to u,

⇒ $-sin(u) = \frac{dt}{du}$

⇒ $sin(u) du = -dt$

Substituting u in terms of t :

⇒ $\int\ {\frac{1}{at^{2} } } \, (-dt)$

⇒ $\int\ {\frac{-1}{at^{2} } } \, dt$

⇒ $\frac{1}{at} + c$  ( c is constant of integration)

Substituting back value of t in terms of u.

⇒ $\frac{1}{t} + c$  

⇒ $\frac{1}{acos(u)} + c$

⇒ $\frac{sec(u)}{a} + c$

Substituting back value of u in terms of x.

⇒ $\frac{sec(u)}{a} + c$

⇒ $\frac{sec(ax+b)}{a} + c$  

The value of given integration is  $\frac{sec(ax+b)}{a} + c$  

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