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If int tan^(7) x dx = f(x) + C then...

If `int tan^(7) x dx = f(x) + C` then

A

f(x) is a polynomial of degree 8 in tan x

B

f(x) is a polynomial of degree 5 in tan x

C

f(x) `= (1)/(6) tan^(6) x - (1)/(4) tan^(4)x + (1)/(2)sec^(2) x + log | cos x| + C`

D

f(x) is a polynomial of degree 6 in tan x

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The correct Answer is:
To solve the integral \( \int \tan^7 x \, dx \), we will use the reduction formula for integrals of the form \( \int \tan^n x \, dx \). ### Step-by-step Solution: 1. **Set up the integral**: Let \( I = \int \tan^7 x \, dx \). 2. **Apply the reduction formula**: The reduction formula for \( \tan^n x \) when \( n \geq 2 \) is given by: \[ \int \tan^n x \, dx = \frac{1}{n-1} \tan^{n-1} x - \int \tan^{n-2} x \, dx \] For \( n = 7 \): \[ I = \frac{1}{6} \tan^6 x - \int \tan^5 x \, dx \] 3. **Evaluate \( \int \tan^5 x \, dx \)**: Let \( J = \int \tan^5 x \, dx \). Using the reduction formula again for \( n = 5 \): \[ J = \frac{1}{4} \tan^4 x - \int \tan^3 x \, dx \] 4. **Evaluate \( \int \tan^3 x \, dx \)**: Let \( K = \int \tan^3 x \, dx \). Again applying the reduction formula for \( n = 3 \): \[ K = \frac{1}{2} \tan^2 x - \int \tan x \, dx \] 5. **Evaluate \( \int \tan x \, dx \)**: The integral of \( \tan x \) is: \[ \int \tan x \, dx = -\log |\cos x| + C \] 6. **Combine results**: Now substitute back: \[ K = \frac{1}{2} \tan^2 x + \log |\cos x| + C \] Substitute \( K \) into \( J \): \[ J = \frac{1}{4} \tan^4 x - \left( \frac{1}{2} \tan^2 x + \log |\cos x| \right) \] Simplifying gives: \[ J = \frac{1}{4} \tan^4 x - \frac{1}{2} \tan^2 x - \log |\cos x| + C \] 7. **Substitute \( J \) into \( I \)**: Substitute \( J \) into \( I \): \[ I = \frac{1}{6} \tan^6 x - \left( \frac{1}{4} \tan^4 x - \frac{1}{2} \tan^2 x - \log |\cos x| \right) \] This simplifies to: \[ I = \frac{1}{6} \tan^6 x - \frac{1}{4} \tan^4 x + \frac{1}{2} \tan^2 x + \log |\cos x| + C \] 8. **Final expression**: Therefore, the result of the integral is: \[ \int \tan^7 x \, dx = \frac{1}{6} \tan^6 x - \frac{1}{4} \tan^4 x + \frac{1}{2} \tan^2 x + \log |\cos x| + C \] ### Conclusion: The correct option from the given choices is: \[ f(x) = \frac{1}{6} \tan^6 x - \frac{1}{4} \tan^4 x + \frac{1}{2} \sec^2 x + \log |\cos x| + C \]
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MCGROW HILL PUBLICATION-INDEFINITE INTEGRATION-SOLVED EXAMPLE ( LEVEL 1 ( SINGLE CORRECT ANSWER TYPE QUESTION ))
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  2. If the primitive of f(x) = (1)/(3 sin x + sin^(3) x) is equal to f(x) ...

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  3. The value of int (cos x)/(sin x + cos x) dx is

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  4. If the primitive of (1)/(e^(x)-1)^(2) is f(x) - log |g (x)| + C then

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  5. If (dx)/(sin^(4)x+cos^(4)x)=(1)/(sqrt(2))tan^(-1)f(x)+C

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  6. The 1int(sin^(2)x)/(cos^(6)x)dx is a

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  7. If int tan^(7) x dx = f(x) + C then

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  8. If int(dx)/(sin x cos x)=log|f(x)|+C then

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  9. If polynomials P and Q satisfyint[(3x-1)cosx+(1-2x)sinx ]dx=P cosx+Qsi...

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  10. Let f(x) = inte^(x^(2))(x-2)(x-3)(x-4)dx then f increases on

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  11. int (x^2 -1 )/ (x^3 sqrt(2x^4 - 2x^2 +1))dx is equal to

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  12. Let I =int(e^(x))/(e^(4x)+e^(2x)+1)dx , J = int(e^(-x))/(e^(-4x)+e^(-...

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  13. The value of sqrt(2)int(sinx)/(sin(x-(pi)/(4)))dx , is

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  14. If int[log(logx)+(1)/((logx)^(2))]dx =x[f(x) - g(x)] + C, then :

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  15. If f((3x-4)/(3x+4))=x+2, then int f(x)dx is equal to

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  16. intcos{2tan^(-1)sqrt((1-x)/(1+x))}dx is equal to

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  17. If u =-f''(theta) sintheta + f'(theta) costheta and v = f''(theta) cos...

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