Advertisements
Advertisements
Question
\[\int \sin^3 x \cos^6 x \text{ dx }\]
Sum
Solution
∫ sin3 x . cos6 x dx
= ∫ sin2 x . cos6 x . sin x dx
= ∫ (1 – cos2 x) . cos6 x . sin x dx
Let cos x = t
⇒ –sin x dx = dt
Now, ∫ (1 – cos2 x) . cos6 x . sin x dx
= –∫ (1 – t2) . t6 dt
= ∫ (t2 – 1) t6 dt
= ∫ (t8 – t6) dt
\[= \frac{t^9}{9} - \frac{t^7}{7} + C\]
\[ = \frac{\cos^9 x}{9} - \frac{\cos^7 x}{7} + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\frac{1 - \cos 2x}{1 + \cos 2x} dx\]
\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{x^2 - 2x + 4} dx\]
\[\int\frac{1}{\sqrt{x + a} + \sqrt{x + b}} dx\]
\[\int \text{sin}^2 \left( 2x + 5 \right) \text{dx}\]
\[\int\frac{\left( x + 1 \right) e^x}{\sin^2 \left( \text{x e}^x \right)} dx\]
\[\int \cot^5 x \text{ dx }\]
\[\int\frac{\sin 8x}{\sqrt{9 + \sin^4 4x}} dx\]
\[\int\frac{\cos x}{\sqrt{4 - \sin^2 x}} dx\]
` ∫ {x-3} /{ x^2 + 2x - 4 } dx `
\[\int\frac{x^3}{x^4 + x^2 + 1}dx\]
\[\int\frac{x^2 + x - 1}{x^2 + x - 6}\text{ dx }\]
\[\int\frac{6x - 5}{\sqrt{3 x^2 - 5x + 1}} \text{ dx }\]
\[\int\frac{x}{\sqrt{x^2 + x + 1}} \text{ dx }\]
\[\int\frac{1}{4 \sin^2 x + 5 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{p + q \tan x} \text{ dx }\]
\[\int\frac{8 \cot x + 1}{3 \cot x + 2} \text{ dx }\]
\[\int x^2 \text{ cos x dx }\]
\[\int\frac{\text{ log }\left( x + 2 \right)}{\left( x + 2 \right)^2} \text{ dx }\]
\[\int\cos\sqrt{x}\ dx\]
\[\int\frac{\sin^{- 1} x}{x^2} \text{ dx }\]
\[\int\frac{x^3 \sin^{- 1} x^2}{\sqrt{1 - x^4}} \text{ dx }\]
\[\int e^x \frac{x - 1}{\left( x + 1 \right)^3} \text{ dx }\]
\[\int\frac{x^2}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{x^2 + x - 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{1}{\sin x + \sin 2x} dx\]
\[\int\frac{\left( x - 1 \right)^2}{x^4 + x^2 + 1} \text{ dx}\]
\[\int\frac{1}{x^4 + 3 x^2 + 1} \text{ dx }\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{2x + 3}} \text{ dx }\]
Write the anti-derivative of \[\left( 3\sqrt{x} + \frac{1}{\sqrt{x}} \right) .\]
If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then
\[\int \cos^3 (3x)\ dx\]
\[\int\frac{1}{\text{ cos }\left( x - a \right) \text{ cos }\left( x - b \right)} \text{ dx }\]
\[\int\frac{1}{1 - x - 4 x^2}\text{ dx }\]
\[\int\sqrt{\frac{1 + x}{x}} \text{ dx }\]
\[\int\frac{1}{\left( \sin x - 2 \cos x \right) \left( 2 \sin x + \cos x \right)} \text{ dx }\]
\[\int {cosec}^4 2x\ dx\]
\[\int x^3 \left( \log x \right)^2\text{ dx }\]
\[\int\frac{x^2}{x^2 + 7x + 10} dx\]
