Advertisements
Advertisements
Question
\[\int \cos^7 x \text{ dx } \]
Sum
Advertisements
Solution
∫ cos7 x dx
= ∫ cos6 x . cos x dx
= ∫ (cos2 x)3 cos x dx
= ∫ (1 – sin2 x)3 . cos x dx
Let sin x = t
⇒ cos x dx = dt
Now, ∫ (1 – sin2 x)3.cos x dx
= ∫ (1 – t2)3 dt
= ∫ (1 – t6 – 3t2 + 3t4) dt
\[= \left[ t - \frac{t^7}{7} - \frac{3 t^3}{3} + \frac{3 t^5}{5} \right] + C\]
\[ = \sin x - \frac{1}{7} \sin^7 x - \sin^3 x + \frac{3}{5} \sin^5 x + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\left( 2 - 3x \right) \left( 3 + 2x \right) \left( 1 - 2x \right) dx\]
\[\int \left( 3x + 4 \right)^2 dx\]
\[\int\frac{1}{\left( 7x - 5 \right)^3} + \frac{1}{\sqrt{5x - 4}} dx\]
\[\int\left( x + 2 \right) \sqrt{3x + 5} \text{dx} \]
\[\int\frac{x}{\sqrt{x + a} - \sqrt{x + b}}dx\]
\[\int\text{sin mx }\text{cos nx dx m }\neq n\]
\[\int\frac{1}{\sqrt{1 - \cos 2x}} dx\]
\[\int\frac{- \sin x + 2 \cos x}{2 \sin x + \cos x} dx\]
\[\int\frac{\cos^5 x}{\sin x} dx\]
\[\int\frac{1}{\sqrt{x} + x} \text{ dx }\]
\[\int\frac{e^{3x}}{4 e^{6x} - 9} dx\]
\[\int\frac{3 x^5}{1 + x^{12}} dx\]
\[\int\frac{x}{x^2 + 3x + 2} dx\]
\[\int\frac{x^2 + 1}{x^2 - 5x + 6} dx\]
\[\int\frac{x + 2}{\sqrt{x^2 - 1}} \text{ dx }\]
\[\int\frac{1}{\left( \sin x - 2 \cos x \right)\left( 2 \sin x + \cos x \right)} \text{ dx }\]
\[\int\frac{1}{13 + 3 \cos x + 4 \sin x} dx\]
\[\int\frac{1}{\sin x + \sqrt{3} \cos x} \text{ dx }\]
`int 1/(sin x - sqrt3 cos x) dx`
\[\int\frac{\sqrt{16 + \left( \log x \right)^2}}{x} \text{ dx}\]
\[\int\sqrt{x^2 - 2x} \text{ dx}\]
\[\int\frac{x^2 + x - 1}{x^2 + x - 6} dx\]
\[\int\frac{x^2}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{1}{\left( x + 1 \right)^2 \left( x^2 + 1 \right)} dx\]
\[\int\frac{dx}{\left( x^2 + 1 \right) \left( x^2 + 4 \right)}\]
\[\int\frac{1}{x \left( x^4 - 1 \right)} dx\]
\[\int\frac{1}{\cos x \left( 5 - 4 \sin x \right)} dx\]
\[\int e^x \left( 1 - \cot x + \cot^2 x \right) dx =\]
\[\int\frac{\sin x}{3 + 4 \cos^2 x} dx\]
\[\int\frac{\cos 2x - 1}{\cos 2x + 1} dx =\]
\[\int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx\] is equal to
\[\int\frac{1}{\sqrt{x} + \sqrt{x + 1}} \text{ dx }\]
\[\int\frac{\left( \sin^{- 1} x \right)^3}{\sqrt{1 - x^2}} \text{ dx }\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int\frac{x^3}{\sqrt{x^8 + 4}} \text{ dx }\]
\[\int\frac{1}{\sin^4 x + \cos^4 x} \text{ dx}\]
\[\int\frac{6x + 5}{\sqrt{6 + x - 2 x^2}} \text{ dx}\]
\[\int\frac{1 + x^2}{\sqrt{1 - x^2}} \text{ dx }\]
\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx}\]
