What is \(\smallint {\sin ^3}x\cos x\;dx\) equal to?
Solving the Integral \(\smallint {\sin ^3}x\cos x\;dx\)
Let's find the value of the integral \(\smallint {\sin ^3}x\cos x\;dx\). This type of integral can be solved effectively using the method of substitution. We look for a part of the integrand whose derivative is also present in the integrand.
Applying Substitution Method for \(\smallint {\sin ^3}x\cos x\;dx\)
Consider the integrand \({\sin ^3}x\cos x\). We can see that the derivative of \(\sin x\) is \(\cos x\), which is present in the integrand.
Let's use the substitution method:
- Let \(u = \sin x\).
- Differentiate both sides with respect to \(x\): \(\frac{{du}}{{dx}} = \cos x\).
- Rearrange to get \(du\) in terms of \(dx\): \(du = \cos x\;dx\).
Now, substitute \(u\) and \(du\) into the original integral:
\(\smallint {\sin ^3}x\cos x\;dx = \smallint {u^3}\;du\)
Integrating the Substituted Expression
The integral \(\smallint {u^3}\;du\) is a basic power rule integral. The power rule states that \(\smallint u^n\;du = \frac{{u^{n+1}}}{{n+1}} + C\), where \(n \neq -1\) and \(C\) is the constant of integration.
Applying the power rule with \(n=3\):
\(\smallint {u^3}\;du = \frac{{u^{3+1}}}{{3+1}} + C = \frac{{u^4}}{4} + C\)
Substituting Back to Original Variable
Now, we substitute back \(u = \sin x\) into our result:
\(\frac{{u^4}}{4} + C = \frac{{{{\left( {\sin x} \right)}^4}}}{4} + C = \frac{{{{\sin }^4}x}}{4} + C\)
So, the integral \(\smallint {\sin ^3}x\cos x\;dx\) is equal to \(\frac{{{{\sin }^4}x}}{4} + C\).
Comparing with Given Options
Let's compare our result, \(\frac{{{{\sin }^4}x}}{4} + C\), with the given options:
- \(\cos ^4x + c\)
- \(\sin ^4x + c\)
- \(\frac{{{{\left( {1 - {{\sin }^2}x} \right)}^2}}}{4} + c\)
- \(\frac{{{{\left( {1 - {{\cos }^2}x} \right)}^2}}}{4} + c\)
Our result is \(\frac{{{{\sin }^4}x}}{4} + C\). Let's examine Option 4:
\(\frac{{{{\left( {1 - {{\cos }^2}x} \right)}^2}}}{4} + c\)
Recall the fundamental trigonometric identity: \(\sin^2 x + \cos^2 x = 1\). From this, we get \(\sin^2 x = 1 - \cos^2 x\).
Substitute this into Option 4:
\(\frac{{{{\left( {\sin^2 x} \right)}^2}}}{4} + c = \frac{{{{\sin }^4}x}}{4} + c\)
This matches our calculated result. The constant of integration is typically denoted by \(C\) or \(c\), which are equivalent in this context.
Let's look at Option 2 again, \(\sin ^4x + c\). This option is missing the division by 4.
Option 3 is \(\frac{{{{\left( {1 - {{\sin }^2}x} \right)}^2}}}{4} + c\). Using the identity, \(1 - \sin^2 x = \cos^2 x\). Substituting this into Option 3 gives \(\frac{{{{\left( {\cos^2 x} \right)}^2}}}{4} + c = \frac{{{{\cos }^4}x}}{4} + c\), which does not match our result.
Therefore, Option 4 is the correct equivalent form of our integral result.
Step-by-Step Solution Summary
- Identify the integral: \(\smallint {\sin ^3}x\cos x\;dx\).
- Choose the substitution \(u = \sin x\).
- Calculate the differential \(du = \cos x\;dx\).
- Substitute \(u\) and \(du\) into the integral: \(\smallint u^3\;du\).
- Integrate using the power rule: \(\frac{u^4}{4} + C\).
- Substitute back \(u = \sin x\): \(\frac{{{{\sin }^4}x}}{4} + C\).
- Rewrite the result using trigonometric identities to match the options: \(\sin^4 x = (\sin^2 x)^2 = (1 - \cos^2 x)^2\).
- Final result matching an option: \(\frac{{{{\left( {1 - {{\cos }^2}x \right)}^2}}}{4} + C\).
Revision Table: Integral \(\smallint {\sin ^3}x\cos x\;dx\)
Here is a summary of the key steps involved in solving this integration problem.
| Step | Action | Result |
|---|---|---|
| 1 | Original Integral | \(\smallint {\sin ^3}x\cos x\;dx\) |
| 2 | Substitution | \(u = \sin x\) |
| 3 | Differential | \(du = \cos x\;dx\) |
| 4 | Integral in terms of \(u\) | \(\smallint u^3\;du\) |
| 5 | Integration | \(\frac{u^4}{4} + C\) |
| 6 | Substitute back \(x\) | \(\frac{{{{\sin }^4}x}}{4} + C\) |
| 7 | Apply Identity \(\sin^2 x = 1 - \cos^2 x\) | \(\frac{{{{\left( {1 - {{\cos }^2}x \right)}^2}}}{4} + C\) |
Additional Information on Substitution in Integration
The substitution method is a fundamental technique for evaluating integrals. It is essentially the reverse of the chain rule for differentiation.
- When to use substitution: It is often useful when the integrand contains a function and its derivative. For example, \(\smallint f(g(x))g'(x)\;dx\).
- The process:
- Choose a substitution \(u = g(x)\). Often, \(u\) is the 'inner' function or a part of the integrand whose derivative is also present.
- Find the differential \(du\) by differentiating \(u\) with respect to \(x\) and multiplying by \(dx\): \(du = g'(x)\;dx\).
- Rewrite the entire integral in terms of \(u\). Make sure there are no \(x\) terms left.
- Evaluate the new integral with respect to \(u\).
- Substitute back \(g(x)\) for \(u\) to express the result in terms of the original variable \(x\).
- Don't forget the constant of integration, \(C\).
The choice of substitution is crucial. Practice helps in identifying the appropriate substitution quickly.
Similar Questions
What is ∫ (e log x + sin x) cos x dx equal to?
What is \(\rm \int \frac{dx}{\sec x+\tan x}\) equal to?
What is \(\rm \int e^{\left(2\ln x+\ln x^2\right)}dx\) equal to?
What is \(\int \dfrac{dx}{sec^2({tan}^{-1}x)}\) equal to?
- What is ∫(x x ) 2 (1 + ln x)dx equal to ?
What is \(\int (\sin x)^{-1/2} (\cos x)^{-3/2}dx\) equal to?
If \(I_1 =\int\frac{e^x dx}{e^x + e^{-x}}\) and \(I_2 =\int\frac{dx}{e^{2x} + 1},\) then what is I 1+ I 2equal to?
What is \(\smallint \frac{{dx}}{{2{x^2} - 2x + 1}}\) equal to?
What is \(\smallint \frac{{dx}}{{x{{\left( {1 + In\;x} \right)}^n}}}\) equal to (n ≠ 1) ?
What is the value of I 1?
Important Questions from Indefinite Integrals
What is ∫ (e log x + sin x) cos x dx equal to?
Evaluation of \(\displaystyle\int{\dfrac{1-\tan x}{1+\tan x}}\ dx\) is:
The value of \(\int\limits_{\pi /4}^{3\pi /4} {\frac{{dx}}{{1 + \cos x}}} \) is:
The value of \(\int_0^\pi {{{\sin }^6}} x \cdot {\cos ^5}x dx\) is:
- \(\rm \int\frac{2x+3}{x^3+x^2-2x}dx=?\)
