Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
∫ sin3 x . cos5 x dx
= ∫ sin2 x . cos5 x . sin x dx
= ∫ (1 – cos2 x) . cos5 x sin x dx
Let cos x = t
⇒ – sin x dx = dt
⇒ sin x dx = – dt
Now, ∫ (1 – cos2 x) . cos5 x sin x dx
= –∫ (1 – t2) t5 dt
= –∫ (t5 – t7) dt
= ∫(t7 – t5) dt
\[= \frac{t^8}{8} - \frac{t^6}{6} + C\]
\[ = \frac{\cos^8 x}{8} - \frac{\cos^6 x}{6} + C\]
APPEARS IN
संबंधित प्रश्न
\[\int\sqrt{x}\left( 3 - 5x \right) dx\]
` ∫ {sec x "cosec " x}/{log ( tan x) }` 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
\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], then
\[\int\frac{1}{\sqrt{x} + \sqrt{x + 1}} \text{ dx }\]
\[\int\frac{1 + \sin x}{\sin x \left( 1 + \cos x \right)} \text{ dx }\]
Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .
Find : \[\int\frac{e^x}{\left( 2 + e^x \right)\left( 4 + e^{2x} \right)}dx.\]
