^6x ^8x dx= 1. 7x+C 2. ^7x 7 +C 3. 7x 7 +C 4. ^7x+C - Vidyadip

Question

$\int\frac{\sin^6\text{x}}{\cos^8\text{x}}\text{ dx}=$

$\tan7\text{x}+\text{C}$
$\frac{\tan^7\text{x}}{7}+\text{C}$
$\frac{\tan7\text{x}}{7}+\text{C}$
$\sec^7\text{x}+\text{C}$

✓

Answer

$\frac{\tan^7\text{x}}{7}+\text{C}$

Solution:

$\text{I}=\int\frac{\sin^6\text{x dx}}{\cos^8\text{x}}$

$\text{I}=\int\tan^6\text{x}\sec^2\text{x dx}$

Put $\tan\text{x}=\text{t}$

$\sec^2\text{x dx}=\text{dt}$

$\text{I}=\int\text{t}^6\text{dt}$

$\text{I}=\frac{\text{t}^7}{7}+\text{C}$

$\text{I}=\frac{\tan^7\text{x}}{7}+\text{C}$

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