Advertisements
Advertisements
प्रश्न
` ∫ tan^5 x dx `
Advertisements
उत्तर
∫ tan5 x dx
= ∫ tan4 x. tan x dx
= ∫(sec2 x – 1)2 . tan x dx
= ∫ (sec4 x – 2 sec2 x + 1) tan x dx
= ∫ tan x . sec4 x dx – 2 ∫ sec2 x . tan x dx+ ∫ tan x dx
= ∫ sec2 x. sec2 x . tan x dx – 2 ∫ tan x sec2 x dx + ∫ tan x dx
= ∫ (1 + tan2 x) . tan x . sec2 x dx – 2 ∫ tan x . sec2 x dx + ∫ tan x dx
Let I1=∫ (1 + tan2 x) . tan x . sec2 x dx – 2 ∫ tan x . sec2 x dx
And I2=∫ tan x dx
∫ tan5 x dx=I1 + I2
Now, I1=∫ (1 + tan2 x) . tan x . sec2 x dx – 2 ∫ tan x . sec2 x dx
Let tan x = t
⇒ sec2x dx = dt
I1=∫ (1 + tan2 x) . tan x . sec2 x dx – 2 ∫ tan x . sec2 x dx
∫ (1 + t2) . t. dt – 2 ∫ t. dt
∫ (t + t3) dt – 2 ∫ t dt
\[\frac{t^2}{2} + \frac{t^4}{4} - \frac{2 t^2}{2} + C_1 \]
\[ = \frac{t^4}{4} - \frac{t^2}{2} + C_1 \]
\[ = \frac{\tan^4 x}{4} - \frac{\tan^2 x}{2} + C_1\]
APPEARS IN
संबंधित प्रश्न
Integrate the following integrals:
\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]
If \[\int\frac{1}{5 + 4 \sin x} dx = A \tan^{- 1} \left( B \tan\frac{x}{2} + \frac{4}{3} \right) + C,\] then
\[\int\sqrt{\frac{1 - x}{x}} \text{ dx}\]
Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .
