Question

Integral sec²x(sec x. tanx) dx​

Answer 1

GIVEN :–

• A Function –

$\\ \implies\bf \sec^{2} (x) \{ \sec(x). \tan(x) \}$

TO FIND :–

• Integration = ?

SOLUTION :–

• Let –

$\\ \implies\bf I = \int\sec^{2} (x) \{ \sec(x). \tan(x) \}.dx$

• Now let's put –

$\\ \implies\bf \sec(x) = t$

• And –

$\\ \implies\bf \{\sec(x) \tan(x) \}dx= dt$

• So that –

$\\ \implies\bf I = \int (t)^{2}.dt$

• We know that –

$\\ \implies \large{ \pink{{ \boxed{\bf\int x^{n}.dx = \dfrac{ {x}^{n + 1} }{n + 1}}}}}$

• So that –

$\\ \implies\bf I = \dfrac{ {t}^{2 + 1} }{2 + 1} + c$

$\\ \implies\bf I = \dfrac{ {t}^{3} }{3} + c$

$\\ \implies\bf I = \dfrac{1}{3}{t}^{3}+ c$

• Now replace 't' –

$\\ \implies \large \pink{ \boxed{\bf I = \dfrac{1}{3}{\sec}^{3}(x)+ c}}$

Answer 2

I=∫(secx+tanx)29sec2xdx

Let secx+tanx=t

⇒secx−tanx=t1

Now (secxtanx+sec2x)dx=dt

secx(secx+tanx)dx=dt

secxdx=tdt,21(t+t1)=secx

I=21∫t29(t+t1)tdt

=21∫(t2−9+t2−13)dt

=21⎣⎢⎢⎡2−9+1t2−9+1+2−13+1t2−13+1⎦⎥⎥⎤

=−71t271−111t2111

=t211−1(111+7t2)

=−(secx+tanx)2111{111+71(secx+tanx)2}+k